nLab salamander lemma

Context

Homological algebra

homological algebra

and

nonabelian homological algebra

diagram chasing

Contents

Idea

The salamander lemma is a fundamental lemma in homological algebra providing information on the relation between homology groups at different positions in a double complex.

By a simple consequence illustrated in remark 3 below, all the standard diagram chasing lemmas of homological algebra are direct and transparent consequences of this lemma, such as the 3x3 lemma, the four lemma, hence the five lemma, the snake lemma, and the long exact sequence in cohomology corresponding to a short exact sequence.

These lemmas are old and classical, but their traditional proofs are, while elementary, not very illuminating. The Salamander lemma serves to make the mechanism behind these lemmas more transparent and also make evident a host of further lemmas of this kind not traditionally considered, such as an nxn lemma for all $n\in ℕ$.

The Salamander lemma

The Salamander lemma, prop. 1 below, is a statement about the exactness of a sequence naturally associated with any morphism in a double complex. In the Preliminaries we first introduce this sequence itself.

Preliminaries

Remark

(general assumption/convention)

As always in homological algebra, when we consider elements of objects in the abelian category $𝒜$ it is either assumed that $𝒜$ is of the form $R$Mod for some ring $R$, or that one of the Embedding theorems has been used to embed it into such, by a faithful and exact functor, and these elements are actual elements of the sets underlying these modules. In the following many of the proofs are spelled out in terms of elements this way and we will not always repeat this assumption. This should help to amplify how utterly elementary the salamander lemma is. But explicitly element-free/general abstract proofs can of course be given with not much more effort, too, see (Wise).

Definition

Let ${A}_{••}$ a double complex in some abelian category $𝒜$, hence a chain complex of chain complexes ${A}_{•,•}\in {\mathrm{Ch}}_{•}\left({\mathrm{Ch}}_{•}\left(𝒜\right)\right)$, hence a diagram of the form

$\begin{array}{ccccc}& & ⋮& & ⋮\\ & & {↓}^{{\partial }^{\mathrm{vert}}}& & {↓}^{{\partial }^{\mathrm{vert}}}\\ \cdots & \to & {A}_{n,m}& \stackrel{{\partial }^{\mathrm{hor}}}{\to }& {A}_{n-1,m}& \to & \cdots \\ & & {↓}^{{\partial }^{\mathrm{vert}}}& & {↓}^{{\partial }^{\mathrm{vert}}}\\ \cdots & \to & {A}_{n,m-1}& \stackrel{{\partial }^{\mathrm{hor}}}{\to }& {A}_{n-1,m-1}& \to & \cdots \\ & & {↓}^{{\partial }^{\mathrm{vert}}}& & {↓}^{{\partial }^{\mathrm{vert}}}\\ & & ⋮& & ⋮\end{array}$\array{ && \vdots && \vdots \\ & & \downarrow^{\mathrlap{\partial^{vert}}} && \downarrow^{\mathrlap{\partial^{vert}}} \\ \cdots &\to & A_{n,m} &\stackrel{\partial^{hor}}{\to}& A_{n-1,m} & \to & \cdots \\ & & \downarrow^{\mathrlap{\partial^{vert}}} && \downarrow^{\mathrlap{\partial^{vert}}} \\ \cdots &\to & A_{n,m-1} &\stackrel{\partial^{hor}}{\to}& A_{n-1,m-1} & \to & \cdots \\ & & \downarrow^{\mathrlap{\partial^{vert}}} && \downarrow^{\mathrlap{\partial^{vert}}} \\ && \vdots && \vdots }

where ${\partial }^{\mathrm{hor}}\circ {\partial }^{\mathrm{hor}}=0$, where ${\partial }^{\mathrm{vert}}\circ {\partial }^{\mathrm{vert}}=0$ and where all squares commute, ${\partial }^{\mathrm{hor}}\circ {\partial }^{\mathrm{vert}}={\partial }^{\mathrm{vert}}\circ {\partial }^{\mathrm{hor}}$.

Let $A≔{A}_{kl}$ be any object in the double complex at any position $\left(k,l\right)$. This is the source and target of horizontal, vertical and diagonal (unique composite of a horizontal and a vertical) morphisms to be denoted as follows:

$\begin{array}{cc}{↘}^{{\partial }_{\mathrm{in}}^{\mathrm{diag}}}& {↓}^{{\partial }_{\mathrm{in}}^{\mathrm{vert}}}\\ \stackrel{{\partial }_{\mathrm{in}}^{\mathrm{hor}}}{\to }& A& \stackrel{{\partial }_{\mathrm{out}}^{\mathrm{hor}}}{\to }\\ & {↓}^{{\partial }_{\mathrm{out}}^{\mathrm{vert}}}& {↘}^{{\partial }_{\mathrm{out}}^{\mathrm{diag}}}\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ \searrow^{\mathrlap{\partial_{in}^{diag}}} & \downarrow^{\mathrlap{\partial_{in}^{vert}}} \\ \stackrel{\partial_{in}^{hor}}{\to} & A & \stackrel{\partial_{out}^{hor}}{\to} \\ & \downarrow^{\mathrlap{\partial_{out}^{vert}}} & \searrow^{\mathrlap{\partial_{out}^{diag}}} } \,.

Define

• ${A}^{\mathrm{hor}}≔\mathrm{ker}\left({\partial }_{\mathrm{out}}^{\mathrm{hor}}\right)/\mathrm{im}\left({\partial }_{\mathrm{in}}^{\mathrm{hor}}\right)\in 𝒜$ – the horizontal chain homology at $X$;

• ${A}^{\mathrm{vert}}≔\mathrm{ker}\left({\partial }_{\mathrm{out}}^{\mathrm{vert}}\right)/\mathrm{im}\left({\partial }_{\mathrm{in}}^{\mathrm{vert}}\right)\in 𝒜$ – the vertical chain homology at $X$;

• ${}^{\square }A≔\frac{\mathrm{ker}\left({\partial }_{\mathrm{out}}^{\mathrm{hor}}\right)\cap \mathrm{ker}\left({\partial }_{\mathrm{out}}^{\mathrm{vert}}\right)}{\mathrm{im}\left({\partial }_{\mathrm{in}}^{\mathrm{diag}}\right)}\in 𝒜$ – the “receptor” at $A$;

• ${A}_{\square }≔\frac{\mathrm{ker}\left({\partial }_{\mathrm{out}}^{\mathrm{diag}}\right)}{\mathrm{im}\left({\partial }_{\mathrm{in}}^{\mathrm{hor}}\right)\oplus \mathrm{im}\left({\partial }_{\mathrm{in}}^{\mathrm{vert}}\right)}$ – the “donor” at $A$;

where $\mathrm{ker}\left(-\right)$ denotes the kernel of a map, $\mathrm{im}\left(-\right)$ the image of a map and $\frac{{N}_{1}}{{N}_{2}}$ the quotient module of the module ${N}_{1}$ by a submodule ${N}_{2}↪{N}_{1}$ and ${N}_{1}\oplus {N}_{2}$ the direct sum of two modules.

Lemma

The identity on representatives in $A$ induces a commuting diagram of homomorphisms from the donor of $A$ to the receptor of $A$, def. 1, via the horizontal and vertical homology groups at $A$:

$\begin{array}{ccc}& & {}^{\square }A\\ & ↙& & ↘\\ {A}^{\mathrm{vert}}& & & & {A}^{\mathrm{hor}}\\ & ↘& & ↙\\ & & {A}_{\square }\phantom{\rule{thinmathspace}{0ex}}.\end{array}$\array{ && {}^\Box A \\ & \swarrow && \searrow \\ A^{vert} &&&& A^{hor} \\ & \searrow && \swarrow \\ && A_\Box \,. }

These morphisms are to be called the intramural maps of $A$.

This is immediate, but here is a way to make it fully explicit:

Proof

The statement that the top two morphisms exist and are given by the identity on representatives is that if $\left[a\right]\in {}^{\square }A$ is represented by $a\in A$, then $a$ also represents an element in ${A}^{\mathrm{hor}}$ and ${A}^{\mathrm{vert}}$. But this is the very definition of ${}^{\square }A$: being a quotient module of $\mathrm{ker}\left({\partial }_{\mathrm{out}}^{\mathrm{vert}}\right)\cap \mathrm{ker}\left({\partial }_{\mathrm{out}}^{\mathrm{hor}}\right)$ means that its elements are represented by elements of $A$ that are annihiliated by both ${\partial }_{\mathrm{out}}^{\mathrm{vert}}$ and ${\partial }_{\mathrm{out}}^{\mathrm{hor}}$. Moreover if $\left[a\right]\in {}^{\square }A$ is 0 then $a$ also represents the 0-element in ${A}^{\mathrm{hor}}$ and in ${A}^{\mathrm{vert}}$, because by definition of ${}^{\square }A$ it is then in the image of ${\partial }^{\mathrm{hor}}\circ {\partial }^{\mathrm{vert}}={\partial }^{\mathrm{vert}}\circ {\partial }^{\mathrm{hor}}$. Moreover it is clear that everything respects addition of module elements and the action by the ring $R$, hence the top morphisms are well defined module homomorphisms.

Similarily the bottom two morphisms exist and are given by the identity on representatives by the very definition of ${A}_{\square }$: this being a quotient of the kernel of ${\partial }^{\mathrm{vert}}{\partial }^{\mathrm{hor}}={\partial }^{\mathrm{hor}}{\partial }^{\mathrm{vert}}$ it contains in particular the elements that are represented in $A$ by elements in the kernel of ${\partial }^{\mathrm{hor}}$ and in the kernel of of ${\partial }^{\mathrm{vert}}$ separately. And if $\left[a\right]$ is the 0-element in ${A}^{\mathrm{hor}}$ or ${A}^{\mathrm{vert}}$ then $a$ is in the image of ${\partial }^{\mathrm{hor}}$ or of ${\partial }^{\mathrm{vert}}$, respectively, and since these two images are quotiented out to obtain ${A}_{\square }$, such $a$ also represent the 0-element there.

Remark

In lemma 1 “intramural” is meant to allude to the fact that these morphisms go between objects that all come from $A$ and hence remain “in the context of $A$”. The “extramural” maps to follow in lemma 2 below instead go from objects associated to some $A$ to objects associated to some $B$, so they go “out of the context of $A$”.

Lemma

For $\partial :A\to B$ is any morphism in the double complex (either horizontal or vertical), when applied to representatives in $A$ it induces a homomorphism

${A}_{\square }\to {}^{\square }B$A_\Box \to {}^\Box B

from the donor of $A$ to the receptor of $B$, def. 1, to be called the extramural map associated with $\partial$.

Again this is immediate, but here is a way to make it explicit:

Proof

We discuss the case that $\partial ={\partial }^{\mathrm{hor}}$ is a horizontal differential. The other case works verbatim the same way, only with the roles of ${\partial }^{\mathrm{hor}}$ and ${\partial }^{\mathrm{vert}}$ interchanged.

By definition 1, an $\left[a\right]\in {A}_{\square }$ is represented by an $a\in A$ for which ${\partial }^{\mathrm{vert}}{\partial }^{\mathrm{hor}}a=0$. The claim is that ${\partial }^{\mathrm{hor}}a$ then represents an element in ${}^{\square }B$ such that this is a module homomorphism.

• We have ${\partial }^{\mathrm{vert}}\left({\partial }^{\mathrm{hor}}a\right)=0$ by assumption on $a$ and ${\partial }^{\mathrm{hor}}\left({\partial }^{\mathrm{hor}}a\right)=0$ by the chain complex property. Hence ${\partial }^{\mathrm{hor}}a$ represents an element in ${}^{\square }B$.

• If $\left[a\right]=0$ then there is $c$ such that $a={\partial }^{\mathrm{hor}}c$ or $d$ such that $a={\partial }^{\mathrm{vert}}d$. In the first case the chain complex property gives that ${\partial }^{\mathrm{hor}}a={\partial }^{\mathrm{hor}}{\partial }^{\mathrm{hor}}c=0$ and hence $\left[{\partial }^{\mathrm{hor}}a\right]=0$ in ${}^{\square }B$, in the second ${\partial }^{\mathrm{hor}}a={\partial }^{\mathrm{hor}}{\partial }^{\mathrm{vert}}d$ which is also 0 in ${}^{\square }B$ since this is the quotient by $\mathrm{im}\left({\partial }^{\mathrm{hor}}{\partial }^{\mathrm{vert}}\right)$.

Remark

(central idea on diagram chasing)

It is useful in computations as those shown below in Implications - The diagram chasing lemmas to draw the extramural morphisms of lemma 2 as follows.

1. For a horizontal ${\partial }^{\mathrm{hor}}:A\to B$ we draw the induced extramural map as

$\begin{array}{cccc}& & & \square \\ A& & ↗& & B\\ & \square \end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ &&& \Box \\ A && \nearrow & & B \\ & \Box } \,.
2. For a vertical ${\partial }^{\mathrm{vert}}:A\to B$ we draw the induced extramural map as

$\begin{array}{cc}& A\\ & & \square \\ & ↙\\ \square \\ & B\end{array}$\array{ & A \\ && \Box \\ & \swarrow \\ \Box \\ & B }

This notation makes it manifest that in every double complex ${X}_{•,•}$ the extramural maps form long diagonal zigzags between donors and receptors

$\begin{array}{ccccccccccc}& & & & & & & & & & ⋰\\ & & & & & & & {X}_{k+1,l}& & ↗\\ & & & & & & & & \square \\ & & & & & & & ↙\\ & & & & & & \square \\ & & & {X}_{k,l+1}& & ↗& & {X}_{k,l}\\ & & & & \square \\ & & & ↙\\ & & \square \\ & ↗& & {X}_{k-1,l+1}\\ ⋰\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ && &&&&&&&& \udots \\ && &&&&& X_{k+1, l} && \nearrow \\ && &&&&&& \Box \\ && &&&&& \swarrow \\ && &&&& \Box \\ && &X_{k,l+1} && \nearrow & & X_{k,l} \\ && && \Box \\ && &\swarrow \\ && \Box \\ & \nearrow & & X_{k-1, l+1} \\ \udots } \,.

But moreover, the intramural maps relate the donors and receptors in particular at the far end of these zigzags back to the actual homology groups of interest:

$\begin{array}{cccccccccccccccc}& & & & & & & & & & & & & & & \square \\ & & & & & & & & & & & & & & ⋰& & B\\ & & & & & & & & & & & {X}_{k+1,l}& & ↗& & & & ↘\\ & & & & & & & & & & & & \square & & & & & & {B}^{\mathrm{hor}}\\ & & & & & & & & & & & ↙\\ & & & & & & & & & & \square \\ & & & & & & & {X}_{k,l+1}& & ↗& & {X}_{k,l}\\ & & & & & & & & \square \\ & & & & & & & ↙\\ {A}^{\mathrm{hor}}& & & & & & \square \\ & ↘& & & & ↗& & {X}_{k-1,l+1}\\ & & A& & ⋰\\ & & & \square \end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ && && && &&&&&&&& & \Box \\ && && && &&&&& && & \udots && B \\ && && && &&&&& X_{k+1, l} && \nearrow & && & \searrow \\ && && && &&&&& & \Box && && & & B^{hor} \\ && & & && &&&&& \swarrow \\ && & & && &&&& \Box \\ && & & && &X_{k,l+1} && \nearrow & & X_{k,l} \\ && & & && && \Box \\ && & & && &\swarrow \\ A^{hor} && & & && \Box \\ &\searrow & & & & \nearrow & & X_{k-1, l+1} \\ && A & & \udots \\ && & \Box } \,.

This means that in order to get “far diagonal identifications” of homology groups in a double complex, all one needs is sufficient conditions that all the intramural and extramural maps in a “long salamander” like this are all isomorphisms.

These turn out to be certain exactness conditions to be checked/imposed locally at each of the positions involved in a long salamander like this discussed in Intramural and extramural isomorphism below. All the long diagonal identifications of the standard diagram chasing lemmas follows by piecing together such long salamanders. This is discussed in the Implications below.

Remark

Morever, it is useful to combine the exramural notation of remark 3 with the evident diagonal notation for the intramural maps, lemma 1, which allow extramural maps to “enter at the receptor” and “exit at the donor” of a given entry in the double complex. Together these two notations yield for every piece of a double complex of the form

$\begin{array}{c}C\\ {↓}^{{\partial }^{\mathrm{vert}}}\\ A& \stackrel{{\partial }^{\mathrm{hor}}}{\to }& B\\ & & {↓}^{{\partial }^{\mathrm{vert}}}\\ & & D\end{array}$\array{ C \\ \downarrow^{\mathrlap{\partial^{vert}}} \\ A &\stackrel{\partial^{hor}}{\to}& B \\ && \downarrow^{\mathrlap{\partial^{vert}}} \\ && D }

the salamander-shaped diagrams of mural maps

$\begin{array}{cc}& C\\ & & \square \\ & ↙\\ \square & & & & \square \\ & A& & ↗& & B\\ & & \square & & & & \square \\ & & & & & ↙\\ & & & & \square \\ & & & & & D\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ & C \\ && \Box \\ & \swarrow \\ \Box &&& & \Box \\ & A && \nearrow & & B \\ && \Box & && & \Box \\ &&&&& \swarrow \\ &&&& \Box \\ &&&&& D } \,.

These give the salamander lemma, prop. 1 below, its name.

Lemma

For ${\partial }^{\mathrm{hor}}:A\to B$ any horizontal morphism in the double complex, the canonically induced morphism ${A}^{\mathrm{vert}}\to {B}^{\mathrm{vert}}$ on vertical homology is the composite of the above intramural and extramural maps:

${A}^{\mathrm{vert}}\to {A}_{\square }\to {}^{\square }B\to {B}^{\mathrm{vert}}\phantom{\rule{thinmathspace}{0ex}}.$A^{vert} \to A_{\Box} \to {}^\Box B \to B^{vert} \,.
Proof

By the above, on representatives the first map is the identity, the second is ${\partial }^{\mathrm{hor}}$ and the third again the identity. Hence the total map is given on representative by ${\partial }^{\mathrm{hor}}$.

Statement

Proposition

(Salamander lemma)

If a diagram

$\begin{array}{c}C\\ {↓}^{{\partial }^{\mathrm{vert}}}\\ A& \stackrel{{\partial }^{\mathrm{hor}}}{\to }& B\\ & & {↓}^{{\partial }^{\mathrm{vert}}}\\ & & D\end{array}$\array{ C \\ \downarrow^{\mathrlap{\partial^{vert}}} \\ A &\stackrel{\partial^{hor}}{\to}& B \\ && \downarrow^{\mathrlap{\partial^{vert}}} \\ && D }

is part of a double complex in an abelian category, then there is a 6-term long exact sequence running horizontally in

$\begin{array}{ccc}& & {}^{\square }A\\ & ↗& & ↘\\ {C}_{\square }& & \to & & {A}^{\mathrm{hor}}& \to & {A}_{\square }& \to & {}^{\square }B& \to & {B}^{\mathrm{hor}}& & \to & & {}^{\square }D\\ & & & & & & & & & & & ↘& & ↗\\ & & & & & & & & & & & & {B}_{\square }\end{array}\phantom{\rule{thinmathspace}{0ex}},$\array{ && {}^\Box A \\ & \nearrow && \searrow \\ C_\Box &&\to&& A^{hor} &\to& A_{\Box} &\to& {}^{\Box} B &\to& B^{hor} &&\to&& {}^{\Box}D \\ && && && && && & \searrow && \nearrow \\ && && && && && && B_{\Box} } \,,

where all the elementary morphisms are the unique intramural maps from lemma 1 and the extramural maps from lemma 2 – they are the morphisms of the salamander diagram of remark 4.

Dually, if a diagram

$\begin{array}{ccc}C& \stackrel{{\partial }^{\mathrm{hor}}}{\to }& A\\ & & {↓}^{{\partial }^{\mathrm{vert}}}\\ & & B& \stackrel{{\partial }^{\mathrm{hor}}}{\to }& D\end{array}$\array{ C &\stackrel{\partial^{hor}}{\to}& A \\ && \downarrow^{\mathrlap{\partial^{vert}}} \\ && B &\stackrel{\partial^{hor}}{\to}& D }

is part of a double complex, then there is a 6-term long exact sequence running horizontally in

$\begin{array}{ccc}& & {}^{\square }A\\ & ↗& & ↘\\ {C}_{\square }& & \to & & {A}^{\mathrm{vert}}& \to & {A}_{\square }& \to & {}^{\square }B& \to & {B}^{\mathrm{vert}}& & \to & & {}^{\square }D\\ & & & & & & & & & & & ↘& & ↗\\ & & & & & & & & & & & & {B}_{\square }\end{array}\phantom{\rule{thinmathspace}{0ex}}$\array{ && {}^\Box A \\ & \nearrow & & \searrow \\ C_\Box &&\to&& A^{vert} &\to& A_{\Box} &\to& {}^\Box B &\to& B^{vert} &&\to&& {}^\Box D \\ && && && && && & \searrow && \nearrow \\ && && && && && && B_\Box } \,

This is (Bergman, lemma 1.7).

Proof

We spell out the proof of the first case. That of the second case is verbatim the same, only with the roles of ${\partial }^{\mathrm{hor}}$ and ${\partial }^{\mathrm{vert}}$ interchanged.

By lemmas 1 and 2, all the maps are given on representatives either by identities or by the differentials of the double complex themselves. Using this we may check exactness at each position explicitly:

1. exactness at ${C}_{\square }\to {A}^{\mathrm{hor}}\to {A}_{\square }$.

An element $\left[a\right]\in {A}^{\mathrm{hor}}$ is in the kernel of ${A}^{\mathrm{hor}}\to {A}_{\square }$ if there is $c$ and $d$ such that $a={\partial }^{\mathrm{vert}}c+{\partial }^{\mathrm{hor}}d$. The $c$ that satisfy this equation hence satisfy ${\partial }^{\mathrm{hor}}{\partial }^{\mathrm{vert}}c=0$, hence represent elements in ${C}_{\square }$ and so the map ${\partial }^{\mathrm{hor}}:{C}_{\square }\to {A}^{\mathrm{hor}}$ hits all of the kernel of ${A}^{\mathrm{hor}}\to {A}_{\square }$. Also it clearly hits at most this kernel.

2. exactness at ${A}^{\mathrm{hor}}\to {A}_{\square }\to {}^{\square }B$.

Suppose $\left[a\right]\in {A}_{\square }$ is in the kernel of ${A}_{\square }\to {}^{\square }B$. This means that there is $c$ such that ${\partial }^{\mathrm{hor}}a={\partial }^{\mathrm{hor}}{\partial }^{\mathrm{vert}}c$, hence that ${\partial }^{\mathrm{hor}}\left(a-{\partial }^{\mathrm{vert}}c\right)=0$. But $\left[a\right]=\left[a-{\partial }^{\mathrm{vert}}c\right]\in {A}_{\square }$ and so this says that $\left[a\right]$ is the image under ${A}^{\mathrm{hor}}\to {A}_{\square }$ of the element represented there by $a-{\partial }^{\mathrm{vert}}c$. Conversely, clearly everything in that image is in the kernel of ${A}_{\square }\to {}^{\square }B$.

3. exactness at ${A}_{\square }\to {}^{\square }B\to {B}^{\mathrm{hor}}$

An element $\left[b\right]\in {}^{\square }B$ is in the kernel of ${}^{\square }B\to {B}^{\mathrm{hor}}$ if there is $a$ such that ${\partial }^{\mathrm{hor}}a=b$. But since the representative $b$ of $\left[b\right]\in {}^{\square }B$ has to satisfy in particular ${\partial }^{\mathrm{hor}}b=0$ it follows that ${\partial }^{\mathrm{hor}}{\partial }^{\mathrm{vert}}a=0$ and hence that $a$ represents an alement in ${A}_{\square }$, hence that $\left[b\right]$ is in the image of ${A}_{\square }\to {}^{\square }B$. Conversely, clearly every element in that image is in the kernel of ${}^{\square }B\to {B}^{\mathrm{hor}}$.

4. exactness at ${}^{\square }B\to {B}^{\mathrm{hor}}\to {}^{\square }D$

An element $\left[b\right]\in {B}^{\mathrm{hor}}$ is in the kernel of ${B}^{\mathrm{hor}}\to {}^{\square }D$ if there is $a$ with ${\partial }^{\mathrm{vert}}b={\partial }^{\mathrm{vert}}{\partial }^{\mathrm{hor}}a$, hence ${\partial }^{\mathrm{vert}}\left(b-{\partial }^{\mathrm{hor}}a\right)=0$ Since in addition ${\partial }^{\mathrm{hor}}\left(b-{\partial }^{\mathrm{hor}}a\right)=0$ by assumption on $b$ and the chain complex property, this says that $\left[b\right]=\left[b+{\partial }^{\mathrm{hor}}a\right]$ is in the image of ${}^{\square }B\to {B}^{\mathrm{hor}}$. Moreover, clearly everything in this image is in the kernel of ${B}^{\mathrm{hor}}\to {}^{\square }D$.

Intramural and extramural isomorphisms

The following two statements are direct concequences (special cases) of the salamander lemma, prop. 1. They give sufficient conditions for the intramural and the extramural maps, lemma 1 and lemma 2, to be isomorphisms. All of of the standard diagram chasing lemmas in homological algebra follow in a natural way from combining these intramural isomorphisms with long zigzags of thse extramural isomorphisms. This is discussed below in The basic diagram chasing lemmas.

Corollary

(extramural isomorphisms)

If the rows of a double complex are exact at the domain and codomain of a horizontal morphism ${\partial }^{\mathrm{hor}}:A\to B$, hence if ${A}^{\mathrm{hor}}=0$ and ${B}^{\mathrm{hor}}=0$, then the extramural map of lemma 2

${A}_{\square }\to {}^{\square }B$A_{\Box} \to {}^\Box B

is an isomorphism.

Similarly if for a vertical morphism $\partial :A\to B$ we have ${A}^{\mathrm{vert}}\simeq 0$ and ${B}^{\mathrm{vert}}\simeq 0$, the induced extramural map

$\begin{array}{c}{A}_{\square }\\ ↓\\ {}^{\square }B\end{array}$\array{ A_\Box \\ \downarrow \\ {}^\Box B }

is an isomorphism.

This appears as (Bergman, cor. 2.1).

It is straightforward to check this directly on elements:

Proof

It is sufficient to show that under the given assumptions both the kernel and the cokernel of the given map are trivial. We discuss the horizontal case. The proof of the vertical case is verbatim the same, only with the roles of ${\partial }^{\mathrm{vert}}$ and ${\partial }^{\mathrm{hor}}$ exchanged.

Suppose an element $\left[a\right]\in {A}_{\mathrm{Box}}$ is in the kernel of ${\partial }^{\mathrm{hor}}:{A}_{\square }\to {}^{\square }B$. By definition of ${}^{\square }B$ this means that there is $c$ such that ${\partial }^{\mathrm{hor}}{\partial }^{\mathrm{vert}}c={\partial }^{\mathrm{hor}}a$, hence such that ${\partial }^{\mathrm{hor}}\left(a-{\partial }^{\mathrm{vert}}c\right)=0$. By assumption that ${A}^{\mathrm{hor}}=0$ this means that there is $d$ such that $a-{\partial }^{\mathrm{vert}}c={\partial }^{\mathrm{hor}}d$. But this means that $a\in \mathrm{im}\left({\partial }^{\mathrm{hor}}\right)\oplus \mathrm{im}\left({\partial }^{\mathrm{vert}}\right)$ and hence $\left[a\right]=0$ in ${A}_{\square }$.

Conversely, consider $\left[b\right]\in {}^{\square }B$. This means that ${\partial }^{\mathrm{vert}}b=0$ and ${\partial }^{\mathrm{hor}}b=0$. By ${B}^{\mathrm{hor}}=0$ the second condition means that there is $a$ such that $b={\partial }^{\mathrm{hor}}a$. Moreover, this $a$ satisfies ${\partial }^{\mathrm{vert}}{\partial }^{\mathrm{hor}}a={\partial }^{\mathrm{vert}}b=0$ by the first condition. Therefore $\left[a\right]\in {A}_{\square }$ and $\left[b\right]$ is its image.

Alternatively, this statement is a direct consequence of the salamander lemma already proven above:

Proof

Under the given assumptions the exact sequence of prop. 1 involves the exact sequence

$0\to {A}_{\square }\to {}^{\square }B\to 0\phantom{\rule{thinmathspace}{0ex}}.$0 \to A_{\Box} \to {}^\Box B \to 0 \,.

This being exact says that the map in the middle has vanishing kernel and cokernel and is hence an isomorphism.

Corollary

(intramural isomorphisms)

In each of the situations in a double complex shown below, if the direction perpendicular to $\partial :A\to B$ or $\partial :B\to A$ is exact at $B$ as indicated in the following, then the two intramural maps from lemma 1, shown in each case on the right, are isomorphisms:

1. $\begin{array}{ccccc}& & ⋮& & ⋮\\ & & ↓& & ↓\\ 0& \to & A& \to & & \to & \cdots \\ & & {↓}^{\partial }& & ↓\\ 0& \to & B& \stackrel{\mathrm{ker}=0}{\to }& & \to & \cdots \\ & & ↓& & ↓\\ & & ⋮& & ⋮\end{array}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⇒\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\begin{array}{ccc}{}^{\square }A& \stackrel{\simeq }{\to }& {A}^{\mathrm{hor}}\\ {A}^{\mathrm{vert}}& \stackrel{\simeq }{\to }& {A}_{\square }\end{array}$\array{ && \vdots && \vdots \\ && \downarrow && \downarrow \\ 0 &\to& A &\to& &\to& \cdots \\ && \downarrow^{\mathrlap{\partial}} && \downarrow \\ 0 &\to& B &\stackrel{ker = 0}{\to}& &\to& \cdots \\ && \downarrow && \downarrow \\ && \vdots && \vdots } \;\;\;\;\;\;\;\;\;\;\;\; \Rightarrow \;\;\;\;\;\;\;\;\;\;\;\; \array{ {}^\Box A &\stackrel{\simeq}{\to}& A^{hor} \\ A^{vert} & \stackrel{\simeq}{\to}& A_{\Box} }
2. $\begin{array}{ccccc}& & 0& & 0\\ & & ↓& & ↓\\ \cdots & \to & A& \stackrel{\partial }{\to }& B& \to & \cdots \\ & & ↓& & {↓}^{\mathrm{ker}=0}\\ \cdots & \to & & \to & & \to & \cdots \\ & & ⋮& & ⋮\end{array}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⇒\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\begin{array}{ccc}{}^{\square }A& \stackrel{\simeq }{\to }& {A}^{\mathrm{vert}}\\ {A}^{\mathrm{hor}}& \stackrel{\simeq }{\to }& {A}_{\square }\end{array}$\array{ && 0 && 0 \\ && \downarrow && \downarrow \\ \cdots &\to& A &\stackrel{\partial}{\to}& B &\to& \cdots \\ && \downarrow && \downarrow^{\mathrlap{ker = 0}} \\ \cdots &\to& &\to& &\to& \cdots \\ && \vdots && \vdots } \;\;\;\;\;\;\;\;\;\;\;\; \Rightarrow \;\;\;\;\;\;\;\;\;\;\;\; \array{ {}^\Box A &\stackrel{\simeq}{\to}& A^{vert} \\ A^{hor} & \stackrel{\simeq}{\to}& A_{\Box} }
3. $\begin{array}{ccccc}& & ⋮& & ⋮\\ & & ↓& & ↓\\ \cdots & \to & & \stackrel{\mathrm{im}=B}{\to }& B& \to & 0\\ & & ↓& & {↓}^{\partial }\\ \cdots & \to & & \to & A& \to & 0\\ & & ↓& & ↓\\ & & ⋮& & ⋮\end{array}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⇒\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\begin{array}{ccc}{A}^{\mathrm{hor}}& \stackrel{\simeq }{\to }& {A}_{\square }\\ {}^{\square }A& \stackrel{\simeq }{\to }& {A}^{\mathrm{vert}}\end{array}$\array{ && \vdots && \vdots \\ && \downarrow && \downarrow \\ \cdots &\to& &\stackrel{im = B}{\to}& B &\to& 0 \\ && \downarrow && \downarrow^{\mathrlap{\partial}} \\ \cdots &\to& &\to& A &\to& 0 \\ && \downarrow && \downarrow \\ && \vdots && \vdots } \;\;\;\;\;\;\;\;\;\;\;\; \Rightarrow \;\;\;\;\;\;\;\;\;\;\;\; \array{ A^{hor} &\stackrel{\simeq}{\to}& A_{\Box} \\ {}^\Box A &\stackrel{\simeq}{\to}& A^{vert} }
4. $\begin{array}{ccccc}& & ⋮& & ⋮\\ \cdots & \to & & \to & & \to & \cdots \\ & & {↓}^{\mathrm{im}=B}& & ↓\\ \cdots & \to & B& \stackrel{\partial }{\to }& A& \to & \cdots \\ & & ↓& & ↓\\ & & 0& & 0\end{array}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⇒\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\begin{array}{ccc}{A}^{\mathrm{vert}}& \stackrel{\simeq }{\to }& {A}_{\square }\\ {}^{\square }A& \stackrel{\simeq }{\to }& {A}^{\mathrm{vert}}\end{array}$\array{ && \vdots && \vdots \\ \cdots &\to& &\to& &\to& \cdots \\ && \downarrow^{\mathrlap{im = B}} && \downarrow \\ \cdots &\to& B &\stackrel{\partial}{\to}& A &\to& \cdots \\ && \downarrow && \downarrow \\ && 0 && 0 } \;\;\;\;\;\;\;\;\;\;\;\; \Rightarrow \;\;\;\;\;\;\;\;\;\;\;\; \array{ A^{vert} &\stackrel{\simeq}{\to}& A_{\Box} \\ {}^\Box A &\stackrel{\simeq}{\to}& A^{vert} }

This appears as (Bergman, cor. 2.2).

Proof

We spell out the proof of the first item. The others work analogously.

Applying cor. 1 to $0\to B$ yields ${}^{\square }B\simeq {0}_{\square }=0$. Therefore the exact sequence of the Salamander lemma 1 corresponding to

$\begin{array}{c}0\\ ↓\\ 0& \to & A\\ & & ↓\\ & & B\end{array}$\array{ 0 \\ \downarrow \\ 0 &\to& A \\ && \downarrow \\ && B }

ends with

$\cdots \to 0\to {}^{\square }A\to {A}^{\mathrm{hor}}\to 0\phantom{\rule{thinmathspace}{0ex}},$\cdots \to 0 \to {}^\Box A\to A^{hor} \to 0 \,,

which implies the first isomorphism. Analogously, the salamander exact sequence associated with

$\begin{array}{ccc}0& \to & A\\ & & ↓\\ & & B& \to & 0\end{array}$\array{ 0 &\to& A \\ && \downarrow \\ && B &\to& 0 }

begins as

$0\to {A}^{\mathrm{vert}}\to {A}_{\square }\to 0\to \cdots \phantom{\rule{thinmathspace}{0ex}}.$0 \to A^{vert} \to A_{\Box} \to 0 \to \cdots \,.

which gives the second isomorphism.

Implications: The basic diagram-chasing lemmas

We derive the basic diagram chasing lemmas from the salamander lemma, or in fact just from repeated application of the intramural/extramural isomorphisms.

The $3×3$ lemma

We derive the sharp 3x3 lemma from the salamander lemma.

Proposition

If in a diagram of the form

$\begin{array}{ccccccc}& & 0& & 0& & 0\\ & & ↓& & ↓& & ↓\\ 0& \to & A\prime & \to & B\prime & \to & C\prime \\ & & ↓& & ↓& & ↓\\ 0& \to & A& \to & B& \to & C\\ & & ↓& & ↓& & ↓\\ 0& \to & A″& \to & B″& \to & C″\end{array}$\array{ && 0 && 0 && 0 \\ && \downarrow && \downarrow && \downarrow \\ 0 &\to& A' &\to& B' &\to& C' \\ && \downarrow && \downarrow && \downarrow \\ 0 &\to& A &\to& B &\to& C \\ && \downarrow && \downarrow && \downarrow \\ 0 &\to& A'' &\to& B'' &\to& C'' }

all columns and the second and third row are exact, then also the first row is exact.

The following proof is that given in (Bergman, lemma 2.3).

Proof

First of all one notices that the diagram is a double complex: by column-exactness the first row includes as subobjects into the second, so the horizontal maps of the first row are restrictions of the differentials of the second and so at least the first row is a chain complex.

We need to show that ${A\prime }^{\mathrm{hor}}\simeq 0$ and ${B\prime }^{\mathrm{hor}}\simeq 0$.

First consider exactness at $A\prime$. The intramural iso, cor. 2 item 1, of

$\begin{array}{c}A\prime \\ {↓}^{\partial }\\ A& \stackrel{\mathrm{ker}=0}{\to }& B\end{array}$\array{ A' \\ \downarrow^{\mathrlap{\partial}} \\ A &\stackrel{ker = 0}{\to}& B }

is ${}^{\square }A\simeq {A}^{\mathrm{hor}}$, and the one of

$\begin{array}{ccc}A\prime & \stackrel{\partial }{\to }& B\prime \\ & & {↓}^{\mathrm{ker}=0}\\ & & B\end{array}$\array{ A' &\stackrel{\partial}{\to}& B' \\ && \downarrow^{\mathrlap{ker = 0}} \\ && B }

according to cor. 2 item 2 is ${}^{\square }A\prime \simeq {A\prime }^{\mathrm{vert}}=0$. Together this gives the desired exactness from the assumtion that ${A\prime }^{\mathrm{vert}}\simeq 0$ (since all the columns are exact by assumption):

${A\prime }^{\mathrm{hor}}\simeq A{\prime }_{\square }\simeq \left({A\prime }^{\mathrm{vert}}\simeq 0\right)\phantom{\rule{thinmathspace}{0ex}}.${A'}^{hor} \simeq A'_\Box \simeq ({A'}^{vert} \simeq 0) \,.

To apply an analogous argument for ${B\prime }^{\mathrm{hor}}$, we combine this kind of identification with the zigzag of intramural maps along the diagonal

$\begin{array}{ccccc}& & & & {B\prime }^{\mathrm{hor}}\\ & & & & & \square \\ & & & & ↙\\ & & & \square \\ {A}^{\mathrm{vert}}& & ↗& & B\\ & \square \end{array}\phantom{\rule{thinmathspace}{0ex}},$\array{ &&&& {B'}^{hor} \\ &&&&& \Box \\ &&&& \swarrow \\ &&& \Box \\ A^{vert} && \nearrow && B \\ & \Box } \,,

which are isos by cor 1. These appear now in the middle of the following chain of isomorphisms

${B\prime }^{\mathrm{hor}}\simeq \left({B\prime }_{\square }\simeq {}^{\square }B\simeq {A}_{\square }\right)\simeq \left({A}^{\mathrm{vert}}\simeq 0\right)\phantom{\rule{thinmathspace}{0ex}},${B'}^{hor} \simeq \left({B'}_{\Box}\simeq {}^{\Box}B \simeq A_{\Box}\right) \simeq \left(A^{vert} \simeq 0\right)\,,

where the first and the last are intramural isos obtained from cor. 2.

Remark

From this argument it is clear that by directly analogous reasoning we obtain ”$n×n$-lemmas” for arbitrary $n$, see prop. 4 below.

In particular we have the following sharp 3x3 lemma.

Proposition

If in a diagram of the form

$\begin{array}{ccccccc}& & 0& & 0& & 0\\ & & ↓& & ↓& & ↓\\ 0& \to & A\prime & \to & B\prime & \to & C\prime & \to & 0\\ & & ↓& & ↓& & ↓\\ 0& \to & A& \to & B& \to & C& \to & 0\\ & & ↓& & ↓& & ↓\\ 0& \to & A″& \to & B″& \to & C″\\ & & ↓\\ & & 0\end{array}$\array{ && 0 && 0 && 0 \\ && \downarrow && \downarrow && \downarrow \\ 0 &\to& A' &\to& B' &\to& C' &\to& 0 \\ && \downarrow && \downarrow && \downarrow \\ 0 &\to& A &\to& B &\to& C &\to& 0 \\ && \downarrow && \downarrow && \downarrow \\ 0 &\to& A'' &\to& B'' &\to& C'' \\ && \downarrow \\ && 0 }

all columns and the second and third row are exact, then also the first row is exact.

Proof

Exactness in $A\prime$ and $B\prime$ is as in prop. 2. For exactness in $C\prime$ we now use the long zigzag of intramural isomorphisms, cor 1.

$\begin{array}{ccccccccc}& & & & & & & & {C\prime }^{\mathrm{hor}}\\ & & & & & & & & & \square \\ & & & & & & & & ↙\\ & & & & & & & \square \\ & & & & B& & ↗& & C\\ & & & & & \square \\ & & & & ↙\\ & & & \square \\ {A″}^{\mathrm{hor}}& & ↗& & B″\\ & \square \end{array}\phantom{\rule{thinmathspace}{0ex}},$\array{ &&&&&&&& {C'}^{hor} \\ &&&&&&&& & \Box \\ &&&&&&&& \swarrow \\ &&&&&&& \Box \\ &&&& B && \nearrow && C \\ &&&&& \Box \\ &&&& \swarrow \\ &&& \Box \\ {A''}^{hor} && \nearrow && B'' \\ & \Box } \,,

So ${C\prime }^{\mathrm{hor}}\simeq {C}_{\square }$ by the intramural iso, then $\dots \simeq A{″}_{\square }$ by this zigzag of extramural isos, and this finally $\cdots \simeq {A″}^{\mathrm{hor}}\simeq 0$ by another intramural iso and by assumption.

The $n×n$ lemma

The proofs of the $3×3$-lemmas above via long diagonal zigzags of extramural isomorphism clearly generalize from double complexes of size $3×3$ to those of arbitrary finite size.

Proposition

If in a diagram of the form

$\begin{array}{ccccccccccc}& & 0& & 0& & 0& & 0& & \\ & & ↓& & ↓& & ↓& & ↓\\ 0& \to & {X}_{n,n}& \to & {X}_{n-1,n}& \to & {X}_{n-2,n}& \to & {X}_{n-3,n}& \to & \cdots \\ & & ↓& & ↓& & ↓& & ↓\\ 0& \to & {X}_{n,n-1}& \to & {X}_{n-1,n-1}& \to & {X}_{n-2,n-1}& \to & {X}_{n-3,n-1}& \to & \cdots \\ & & ↓& & ↓& & ↓& & ↓\\ 0& \to & {X}_{n,n-2}& \to & {X}_{n-1,n-2}& \to & {X}_{n-2,n-2}& \to & {X}_{n-3,n-2}& \to & \cdots \\ & & ↓& & ↓& & ↓& & ↓\\ 0& \to & {X}_{n,n-3}& \to & {X}_{n-1,n-3}& \to & {X}_{n-2,n-3}& \to & {X}_{n-3,n-3}& \to & \cdots \\ & & ↓& & ↓& & ↓& & ↓\end{array}$\array{ && 0 && 0 && 0 && 0 && \\ && \downarrow && \downarrow && \downarrow && \downarrow \\ 0 &\to& X_{n,n} &\to& X_{n-1,n} &\to& X_{n-2,n} &\to& X_{n-3,n} &\to& \cdots \\ && \downarrow && \downarrow && \downarrow && \downarrow \\ 0 &\to& X_{n,n-1} &\to& X_{n-1,n-1} &\to& X_{n-2,n-1} &\to& X_{n-3,n-1} &\to& \cdots \\ && \downarrow && \downarrow && \downarrow && \downarrow \\ 0 &\to& X_{n,n-2} &\to& X_{n-1,n-2} &\to& X_{n-2,n-2} &\to& X_{n-3,n-2} &\to& \cdots \\ && \downarrow && \downarrow && \downarrow && \downarrow \\ 0 &\to& X_{n,n-3} &\to& X_{n-1,n-3} &\to& X_{n-2,n-3} &\to& X_{n-3,n-3} &\to& \cdots \\ && \downarrow && \downarrow && \downarrow && \downarrow }

all rows except possibly the first ${X}_{•,n}$ as well as all columns except possibly the first ${X}_{n,•}$ are exact, then the homology groups of the first row equal those of the first column in that

$\forall k:{X}_{k,0}^{\mathrm{hor}}\simeq {X}_{0,k}^{\mathrm{vert}}\phantom{\rule{thinmathspace}{0ex}}.$\forall k : X_{k,0}^{hor} \simeq X_{0,k}^{vert} \,.

This appears as (Bergman, lemma 2.6).

Proof

The proof proceeds in direct generalization of the proofs of the 3x3 lemma above: the isomorphism for each $k$ is given by the comoposite of two extramural isomorphism that identify the given homology group with a donor or receptor group, respectively, with a long zigzag of extramural isomorphisms.

The four lemma

We prove the strong four lemma from the salamander lemma.

Proposition

Consider a commuting diagram in $𝒜$ of the form

$\begin{array}{ccccccc}A& \to & B& \stackrel{\xi }{\to }& C& \to & D\\ {↓}^{\tau }& & {↓}^{f}& & {↓}^{g}& & {↓}^{\nu }\\ A\prime & \to & B\prime & \stackrel{\eta }{\to }& C\prime & \to & D\prime \end{array}$\array{ A &\to& B &\stackrel{\xi}{\to}& C &\to& D \\ \downarrow^{\mathrlap{\tau}} && \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{g}} && \downarrow^{\mathrlap{\nu}} \\ A' &\to& B' &\stackrel{\eta}{\to}& C' &\to& D' }

where

1. the rows are exact sequences,

2. $\tau$ is an epimorphism,

3. $\nu$ is a monomorphism.

Then

1. $\xi \left(\mathrm{ker}\left(f\right)\right)=\mathrm{ker}\left(g\right)$

2. $\mathrm{im}\left(f\right)={\eta }^{-1}\left(\mathrm{im}\left(g\right)\right)$

and so in particular

1. if $f$ is a monomorphism then so is $g$;

2. if $g$ is an epimorphism then so is $f$.

Proof

By assumption on $\tau$ and $\nu$ we can complete the diagram to a double complex of the form

$\begin{array}{ccccc}& & 0& & 0\\ & & ↓& & ↓\\ & & \mathrm{ker}\left(f\right)& \stackrel{\xi {\mid }_{\mathrm{ker}\left(f\right)}}{\to }& \mathrm{ker}\left(g\right)& \to & 0\\ & & ↓& & ↓& & ↓\\ A& \to & B& \stackrel{\xi }{\to }& C& \to & D\\ {↓}^{\tau }& & {↓}^{f}& & {↓}^{g}& & {↓}^{\nu }\\ A\prime & \to & B\prime & \stackrel{\eta }{\to }& C\prime & \to & D\prime \\ ↓& & ↓& & ↓\\ 0& \to & B\prime /\mathrm{im}\left(f\right)& \to & C\prime /\mathrm{im}\left(g\right)\\ & & ↓& & ↓\\ & & 0& & 0\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ && 0 && 0 \\ && \downarrow && \downarrow \\ && ker(f) &\stackrel{\xi|_{ker(f)}}{\to}& ker(g) &\to& 0 \\ && \downarrow && \downarrow && \downarrow \\ A &\to& B &\stackrel{\xi}{\to}& C &\to& D \\ \downarrow^{\mathrlap{\tau}} && \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{g}} && \downarrow^{\mathrlap{\nu}} \\ A' &\to& B' &\stackrel{\eta}{\to}& C' &\to& D' \\ \downarrow && \downarrow && \downarrow \\ 0 &\to& B'/im(f) &\to& C'/im(g) \\ && \downarrow && \downarrow \\ && 0 && 0 } \,.

such that

1. all columns are exact;

2. the middle two rows are exact.

For the first statement it is now sufficient to show that $\mathrm{ker}\left(g{\right)}^{\mathrm{hor}}\simeq 0$, for that is immediately equivalent to $\xi \left(\mathrm{ker}\left(f\right)\right)=\mathrm{ker}\left(g\right)$.

To see this we use the intramural isomorphism, cor. 2 item 2, to deduce that

$\mathrm{ker}\left(g{\right)}^{\mathrm{hor}}\simeq \mathrm{ker}\left(g{\right)}_{\square }\phantom{\rule{thinmathspace}{0ex}}.$ker(g)^{hor} \simeq ker(g)_{\Box} \,.

Then the long zigzag of extramural isomorphisms, cor. 1, shows that this is isomorphic to the ${}^{\square }0\simeq 0$ in the bottom left corner of the diagram.

The second statement follows dually: it is implied by $\left(B\prime /\mathrm{im}\left(f\right){\right)}^{\mathrm{hor}}\simeq 0$ for that directly implies that ${\eta }^{-1}\left(\mathrm{im}\left(g\right)\right)\simeq \mathrm{im}\left(f\right)$.

Here the intramural ismorphism to use is

$\left(B\prime /\mathrm{im}\left(f\right){\right)}^{\mathrm{hor}}\simeq {}^{\square }\left(B\prime /\mathrm{im}\left(f\right)\right)$(B'/im(f))^{hor} \simeq {}^{\Box}(B'/im(f))

and then the long sequence of zigzags of extramural ismoporphisms identifies this with the ${0}_{\square }\simeq 0$ in the top right corner.

Remark

The four lemma, in turn, directly implies what is known as the five lemma.

The snake lemma

We discuss a proof of the snake lemma from the salamander lemma.

Proposition

If in a commuting diagram of the form

$\begin{array}{ccccccccc}& & {X}_{1}& \to & {X}_{2}& \to & {X}_{3}& \to & 0\\ & & {↓}^{f}& & {↓}^{g}& & {↓}^{h}\\ 0& \to & {Y}_{1}& \to & {Y}_{2}& \to & {Y}_{3}\end{array}$\array{ && X_1 &\to& X_2 &\to& X_3 &\to& 0 \\ && \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{g}} && \downarrow^{\mathrlap{h}} \\ 0 &\to& Y_1 &\to& Y_2 &\to& Y_3 }

both rows are exact, then there is a long exact sequence

$\mathrm{ker}\left(f\right)\to \mathrm{ker}\left(g\right)\to \mathrm{ker}\left(h\right)\stackrel{\partial }{\to }\mathrm{coker}\left(f\right)\to \mathrm{coker}\left(g\right)\to \mathrm{coker}\left(h\right)$ker(f) \to ker(g) \to ker(h) \stackrel{\partial}{\to} coker(f) \to coker(g) \to coker(h)

starting with the kernels of the three vertical maps and ending with their cokernels (with the middle morphism $\delta$ called the “connecting homomorphism”).

Proof

Consider the completion of the given diagram to a double complex:

$\begin{array}{ccccccccc}& & \mathrm{ker}\left(f\right)& \to & \mathrm{ker}\left(g\right)& \to & \mathrm{ker}\left(h\right)& \to & 0\\ & & ↓& & ↓& & ↓& & ↓\\ \mathrm{ker}\left(l\right)& \to & {X}_{1}& \stackrel{l}{\to }& {X}_{2}& \stackrel{p}{\to }& {X}_{3}& \to & 0\\ ↓& & {↓}^{f}& & {↓}^{g}& & {↓}^{h}& & ↓\\ 0& \to & {Y}_{1}& \stackrel{i}{\to }& {Y}_{2}& \stackrel{r}{\to }& {Y}_{3}& \to & \mathrm{coker}\left(r\right)\\ ↓& & ↓& & ↓& & ↓\\ 0& \to & \mathrm{coker}\left(f\right)& \to & \mathrm{coker}\left(g\right)& \to & \mathrm{coker}\left(h\right)& & \end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ && ker(f) &\to& ker(g) &\to& ker(h) &\to& 0 \\ && \downarrow && \downarrow && \downarrow && \downarrow \\ ker(l) &\to& X_1 &\stackrel{l}{\to}& X_2 &\stackrel{p}{\to}& X_3 &\to& 0 \\ \downarrow && \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{g}} && \downarrow^{\mathrlap{h}} && \downarrow \\ 0 &\to& Y_1 &\stackrel{i}{\to}& Y_2 &\stackrel{r}{\to}& Y_3 &\to& coker(r) \\ \downarrow && \downarrow && \downarrow && \downarrow \\ 0 &\to& coker(f) &\to& coker(g) &\to& coker(h) && } \,.

By assumption and construction, here all columns are exact and the rows are exact at the ${X}_{i}$ and at the ${Y}_{i}$.

Now horizontal exactness at $\mathrm{ker}\left(g\right)$ follows from the intramural isomorphism $\mathrm{ker}\left(g{\right)}^{\mathrm{hor}}\simeq \mathrm{ker}\left(f{\right)}_{\square }$, cor. 2, combined with the zigzag of extramural isomorphisms, cor. 1,

$\begin{array}{cccccccc}& & & & & & & \mathrm{ker}\left(g\right)\\ & & & & & & & & \square \\ & & & & & & & ↙\\ & & & & & & \square \\ & {X}_{1}& & & & ↗& & {X}_{2}\\ & & \square \\ & ↙\\ \square \\ & {Y}_{1}\end{array}$\array{ &&&&&&& ker(g) \\ &&&&&&&& \Box \\ &&&&&&& \swarrow \\ & &&&&& \Box \\ & X_1 &&&& \nearrow & & X_2 \\ && \Box \\ & \swarrow \\ \Box \\ & Y_1 }

which give $\cdots \simeq {}^{\square }{Y}_{1}$ and then by another intramural iso $\cdots \simeq {Y}_{1}^{\mathrm{hor}}$, and finally by assumption $\cdots \simeq 0$.

The exactness as $\mathrm{coker}\left(g\right)$ is shown analogously.

Finally, to build the connecting homomorphism $\mathrm{ker}\left(h\right)\to \mathrm{coker}\left(f\right)$ is the same as giving an isomorphism from $\mathrm{coker}\left(\mathrm{ker}\left(g\right)\to \mathrm{ker}\left(h\right)\right)\simeq \mathrm{ker}\left(h{\right)}^{\mathrm{hor}}$ to $\mathrm{ker}\left(\mathrm{coker}\left(f\right)\to \mathrm{coker}\left(g\right)\right)=\mathrm{coker}\left(f{\right)}^{\mathrm{hor}}$. This is in turn given by the intramural isomorphisms $\mathrm{ker}\left(h{\right)}^{\mathrm{hor}}\simeq \mathrm{ker}\left(h{\right)}_{\square }$ and ${}^{\square }\mathrm{coker}\left(f\right)\simeq \mathrm{coker}\left(f{\right)}^{\mathrm{hor}}$, cor. 2 connected by the zigzag of extramural isomorphisms, cor. 1

$\begin{array}{cccccccccc}& & & & & & & & & \mathrm{ker}\left(h\right)\\ & & & & & & & & & & \square \\ & & & & & & & & & ↙\\ & & & & & & & & \square \\ & & & & & {X}_{2}& & ↗& & {X}_{3}\\ & & & & & & \square \\ & & & & & ↙\\ & & & & \square \\ & {Y}_{1}& & ↗& & {Y}_{2}\\ & & \square \\ & ↙\\ \square \\ & \mathrm{coker}\left(f\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ &&&&&&&&& ker(h) \\ &&&&&&&&&& \Box \\ &&&&&&&&& \swarrow \\ &&&&& && & \Box \\ &&&&& X_2 && \nearrow && X_3 \\ &&&&&& \Box \\ &&&&& \swarrow \\ &&&& \Box \\ & Y_1 && \nearrow & & Y_2 \\ && \Box \\ & \swarrow \\ \Box \\ & coker(f) } \,.
Remark

From the snake lemma one obtains in turn the connecting homomorphism, see there for details.

References

The salamander lemma is due to

based on an earlier unpublished preprint which was circulated (pdf).

An exposition of this is in

A purely category-theoretic proof is in section 2 of

Revised on June 4, 2013 19:12:34 by can I even register here? (129.215.5.255)