# nLab dependent product

Dependent products

category theory

## Applications

#### Limits and colimits

limits and colimits

# Dependent products

## Idea

The dependent product is a universal construction in category theory. It generalizes the internal hom to the situation where the codomain may depend on the domain, and internalizes indexed products. Hence it forms sections of a bundle.

The dual concept is that of dependent sum.

The concept of cartesian product of sets makes sense for any family of sets, while the category-theoretic product makes sense for any family of objects. In each case, however, the family is indexed by a set; how can we get a purely category-theoretic product indexed by an object?

First we need to describe a family of objects indexed by an object; it's common to interpret this as a bundle, that is an arbitrary morphism $\pi: E \to A$. (In Set, $A$ would be the index set of the family, and the fiber of the bundle over an element $x$ of $A$ would be the set indexed by $x$. Conversely, given a family of sets, $E$ can be constructed as its disjoint union.)

In these terms, the cartesian product of the family of sets is the set $S$ of (global) sections of the bundle. This set comes equipped with an evaluation map $ev: S \times A \to E$ such that

$S \times A \stackrel{ev}\to E \stackrel{\pi}\to A$

equals the usual product projection from $S \times A$ to $A$, so $ev$ and $\pi$ are both morphisms in the over category $Set/A$. The universal property of $S$ is that, given any set $T$ and morphism $T \times A \to E$ in $Set/A$, there's a unique map $T \to S$ that makes everything commute.

In other words, $S$ and $ev$ define an adjunction from $Set$ to $Set/A$ in which taking the product with $A$ is the left adjoint and applying this universal property is the right adjoint. This is the basis for the definition below, but we add one further level of generality: We realise that $Set$ is secretly $Set/*$, where $*$ is the one-point set (the final object), and move everything from $Set$ to an arbitrary over category $\mathcal{C}/I$.

## Definitions

Let $\mathcal{C}$ be a category, and $g\colon B \to A$ a morphism in $\mathcal{C}$, such that pullbacks along this morphism exist. These then constitute the base change functor between the corresponding slice categories

$g^* \colon \mathcal{C}_{/A} \to \mathcal{C}_{/B} \,.$
###### Definition

The dependent product along $g$ is, if it exists, the right adjoint functor $\prod_g \colon \mathcal{C}_{/B} \to \mathcal{C}_{/A}$ to the base change along $g$

$(g^* \dashv \prod_g) \colon \mathcal{C}_{/B} \stackrel{\overset{g^* }{\leftarrow}}{\underset{\prod_g}{\to}} \mathcal{C}_{/A} \,.$

So a category with all dependent products is necessarily a category with all pullbacks.

## Properties

### Relation to spaces of sections

###### Proposition

Let $\mathcal{C}$ be a cartesian closed category with all limits and note that $\mathcal{C}_{/\ast}\cong\mathcal{C}$. Let $X \in C$ be any object and identify it with the terminal morphism $X\to *$. Then the dependent product functor

$\mathcal{C}_{/X} \underoverset {\underset{\prod_{x \in X}}{\longrightarrow}} {\overset{- \times X}{\longleftarrow}} {\bot} \mathcal{C}$

sends bundles $P \to X$ to their object of sections.

$\prod_{x \in X} P_x \simeq \Gamma_X(P) := [X,P] \times_{[X,X]} \{id\} \,.$
###### Proof

We check the characterizing natural isomorphism for the adjunction: For every $A \in \mathcal{C}$ we have the following sequence of natural isos:

\begin{aligned} \mathcal{C}_{/X}(A \times X, P \to X) &= \mathcal{C}(A \times X, P ) \times_{\mathcal{C}(A \times X, X)} \{p_2\} \\ & = \mathcal{C}(A, [X,P]) \times_{\mathcal{C}(A,[X,X])} \{\tilde p_2\} \\ &= \mathcal{C}(A, [X,P] \times_{[X,X]} \{Id\}) \\ &= \mathcal{C}(A, \Gamma_X(P)) \end{aligned} \,.
###### Remark

This statement and its proof remain valid in homotopy theory. More in detail, if $\mathcal{C}$ is a simplicial model category, $X$, $A$ and $X \times A$ are cofibrant, $P$ and $X$ are fibrant and $P \to X$ is a fibration, then $\Gamma_X(A)$ as above is the homotopy-correct derived section space.

### Relation to exponential objects / internal homs

As a special case of prop. one obtains exponential objects/internal homs.

Let $\mathcal{C}$ have a terminal object $* \in \mathcal{C}$. Let $A$ and $X$ in $\mathcal{C}$ be objects and let $A \colon A \to *$ and $X \colon X \to *$ be the terminal morphisms.

###### Corollary

The dependent product along $X$ of the arrow obtained by base change of $A$ along $X$ is the exponential object $[X,A]$:

$\prod_{X} X^* A \simeq [X,A] \in \mathcal{C} \,.$
###### Remark

This is essentially a categorified version of the familiar fact that the exponentiation $m^n$ of two natural numbers can be identified with the product $\overset{n}{\overbrace{m\cdot\dots \cdot m}}$ of $n$ copies of $m$.

###### Example

Consider the chain of equivalences from $[X,A]$ to $\prod_{X} X^* A$ in Set: Firstly, the exponential object $[X,A]$ is characterized in $[Y,[X,A]]$ as right adjoint to $[Y\times X,A]$. Secondly, the elements $\theta$ of $[Y\times X,A]$ are in turn in bijection with those functions $(y,x)\mapsto (\theta(y,x),x)$ from $[Y\times X,A\times X]$, that leave the second component fixed. The condition just stated is the definition of arrows in the overcategory ${Set}/X$, between the right projections out of $Y\times X$ resp. $A\times X$. If we identify objects $Z\in{Set}$ with their terminal morphisms $Z:Z\to *$ in ${Set}/*$, those two right projections are the pullbacks $X^* Y$ and $X^* A$, respectively. Thirdly, thus, the subset of $[Y\times X,A\times X]$ we are interested in corresponds to $[X^* Y,X^* A]$ in ${Set}/X$. Finally, the right adjoint to $X^*$ is a functor $\prod_{X}$ from ${Set}/X$ to ${Set}/*$, such that $[X^* Y,X^* A]\simeq [Y, \prod_{X} X^* A]$. Hence $\prod_{X} X^* A$ must correspond to $[X,A]$.

### Relation to type theory and quantification in logic

The dependent product is the categorical semantics of what in type theory is the formation of dependent product types. Under propositions as types this corresponds to universal quantification.

type theorycategory theory
syntaxsemantics
natural deductionuniversal construction
dependent product typedependent product
type formation$\displaystyle\frac{\vdash\: X \colon Type \;\;\;\;\; x \colon X \;\vdash\; A(x)\colon Type}{\vdash \; \left(\prod_{x \colon X} A\left(x\right)\right) \colon Type}$$\left( \array{ A^{\phantom{p_1}} \\ \downarrow^{p_1} \\ X_{\phantom{p_1}}} \in \mathcal{C}\right) \Rightarrow \left( \prod_{x\colon X} A\left(x\right) \in \mathcal{C}\right)$
term introduction$\displaystyle\frac{x \colon X \;\vdash\; a\left(x\right) \colon A\left(x\right)}{\vdash (x \mapsto a\left(x\right)) \colon \prod_{x' \colon X} A\left(x'\right) }$
term elimination$\displaystyle\frac{\vdash\; f \colon \left(\prod_{x \colon X} A\left(x\right)\right)\;\;\;\; \vdash \; x \colon X}{x \colon X\;\vdash\; f(x) \colon A(x)}$
computation rule$(y \mapsto a(y))(x) = a(x)$

## Examples

### In toposes

Dependent products (and sums) exist in any topos:

###### Proposition

For $C$ a topos and $f : A \to I$ any morphism in $C$, both the left adjoint $\sum_f : C/A \to C/I$ as well as the right adjoint $\prod_f: C/A \to C/I$ to $f^*: C/I \to C/A$ exist.

Moreover, $f^*$ preserves the subobject classifier and internal homs.

This is (MacLaneMoerdijk, theorem 2 in section IV, 7).

The dependent product plays a role in the definition of universe in a topos.

### Along $\mathbf{B}H \to \mathbf{B}G$

For $\mathbf{H}$ an (∞,1)-topos and $G$ an group object in $\mathbf{H}$ (an ∞-group), then the slice (∞,1)-topos over its delooping may be identified with the (∞,1)-category of $G$-∞-actions (see there for more):

$Act_G(\mathbf{H}) \simeq \mathbf{H}_{/\mathbf{B}G} \,.$

Under this identification, then dependent product along a morphism of the form $\mathbf{B}H \to \mathbf{B}G$ (corresponding to an ∞-group homomorphism $H \to G$) corresponds to forming coinduced representations.

### Along $\ast \to \mathbf{B}G$

As the special case of the above for $H = 1$ the trivial group we obtain the following:

###### Proposition

Let $\mathbf{H}$ be any (∞,1)-topos and let $G$ be a group object in $\mathbf{H}$ (an ∞-group). Then the dependent product along the canonical point inclusion

$i \;\colon\; \ast \to \mathbf{B}G$

into the delooping of $G$ takes the following form:

There is a pair of adjoint ∞-functors of the form

$\mathbf{H} \underoverset { \underset{\underset{i}{\prod} \simeq [G,-]/G}{\longrightarrow}} { \overset{i^\ast \simeq hofib}{\longleftarrow}} {\bot} \mathbf{H}_{/\mathbf{B}G} \,,$

where

• $hofib$ denotes the operation of taking the homotopy fiber of a map to $\mathbf{B}G$ over the canonical basepoint;

• $[G,-]$ denotes the internal hom in $\mathbf{H}$;

• $[G,-]/G$ denotes the homotopy quotient by the conjugation ∞-action for $G$ equipped with its canonical ∞-action by left multiplication and the argument regarded as equipped with its trivial $G$-$\infty$-action

(for $G = S^1$ then this is the cyclic loop space construction).

Hence for

then there is a natural equivalence

$\underset{ \text{original} \atop \text{fluxes} }{ \underbrace{ \mathbf{H}(\hat X\;,\; A) } } \;\; \underoverset {\underset{oxidation}{\longleftarrow}} {\overset{reduction}{\longrightarrow}} {\simeq} \;\; \underset{ \text{doubly} \atop { \text{dimensionally reduced} \atop \text{fluxes} } }{ \underbrace{ \mathbf{H}(X \;,\; [G,A]/G) } }$

given by

$\left( \hat X \longrightarrow A \right) \;\;\; \leftrightarrow \;\;\; \left( \array{ X && \longrightarrow && [G,A]/G \\ & \searrow && \swarrow \\ && \mathbf{B}G } \right)$
###### Proof

The statement that $i^\ast \simeq hofib$ follows immediately by the definitions. What we need to show is that the dependent product along $i$ is given as claimed.

To that end, first observe that the conjugation action on $[G,X]$ is the internal hom in the (∞,1)-category of $G$-∞-actions $Act_G(\mathbf{H})$. Under the equivalence of (∞,1)-categories

$Act_G(\mathbf{H}) \simeq \mathbf{H}_{/\mathbf{B}G}$

(from NSS 12) then $G$ with its canonical ∞-action is $(\ast \to \mathbf{B}G)$ and $X$ with the trivial action is $(X \times \mathbf{B}G \to \mathbf{B}G)$.

Hence

$[G,X]/G \simeq [\ast, X \times \mathbf{B}G]_{\mathbf{B}G} \;\;\;\;\; \in \mathbf{H}_{/\mathbf{B}G} \,.$

So far this is the very definition of what $[G,X]/G \in \mathbf{H}_{/\mathbf{B}G}$ is to mean in the first place.

But now since the slice (∞,1)-topos $\mathbf{H}_{/\mathbf{B}G}$ is itself cartesian closed, via

$E \times_{\mathbf{B}G}(-) \;\;\; \dashv \;\;\; [E,-]_{\mathbf{B}G}$

it is immediate that there is the following sequence of natural equivalences

\begin{aligned} \mathbf{H}_{/\mathbf{B}G}(Y, [G,X]/G) & \simeq \mathbf{H}_{/\mathbf{B}G}(Y, [\ast, X \times \mathbf{B}G]_{\mathbf{B}G}) \\ & \simeq \mathbf{H}_{/\mathbf{B}G}( Y \times_{\mathbf{B}G} \ast, \underset{p^\ast X}{\underbrace{X \times \mathbf{B}G }} ) \\ & \simeq \mathbf{H}( \underset{hofib(Y)}{\underbrace{p_!(Y \times_{\mathbf{B}G} \ast)}}, X ) \\ & \simeq \mathbf{H}(hofib(Y),X) \end{aligned}

Here $p \colon \mathbf{B}G \to \ast$ denotes the terminal morphism and $p_! \dashv p^\ast$ denotes the base change along it.

### Along $V/G \to \mathbf{B}G$

More generally:

###### Proposition

Let $\mathbf{H}$ be an (∞,1)-topos and $G \in Grp(\mathbf{H})$ an ∞-group.

Let moreover $V \in \mathbf{H}$ be an object equipped with a $G$-∞-action $\rho$, equivalently (by the discussion there) a homotopy fiber sequence of the form

$\array{ V \\ \downarrow \\ V/G & \overset{p_\rho}{\longrightarrow}& \mathbf{B}G }$

Then

1. pullback along $p_\rho$ is the operation that assigns to a morphism $c \colon X \to \mathbf{B}G$ the $V$-fiber ∞-bundle which is associated via $\rho$ to the $G$-principal ∞-bundle $P_c$ classified by $c$:

$(p_\rho)^\ast \;\colon\; c \mapsto P_c \times_G V$
2. the right base change along $p_\rho$ is given on objects of the form $X \times (V/G)$ by

$(p_\rho)_\ast \;\colon\; X \times (V/G) \;\mapsto\; [V,X]/G$
###### Proof

The first statement is NSS 12, prop. 4.6.

The second statement follows as in the proof of prop. : Let

$\left( \array{ Y \\ \downarrow^{\mathrlap{c}} \\ \mathbf{B}G } \right) \;\in\; \mathbf{H}_{/\mathbf{B}G}$

be any object, then there is the following sequence of natural equivalences

\begin{aligned} \mathbf{H}_{/\mathbf{B}G}(Y, [V,X]/G) & \simeq \mathbf{H}_{/\mathbf{B}G}(Y, [V/G, X \times \mathbf{B}G]_{\mathbf{B}G}) \\ & \simeq \mathbf{H}_{/\mathbf{B}G}( Y \times_{\mathbf{B}G} (V/G), \underset{p^\ast X}{\underbrace{X \times \mathbf{B}G }} ) \\ & \simeq \mathbf{H}_{/\mathbf{B}G} ( (p_\rho)_!( P_c \times_G (V/G) ), p^\ast X ) \\ & \simeq \mathbf{H}_{/(V/G)} ( P_c \times_G V, (p_\rho)^\ast p^\ast X ) \\ & \simeq \mathbf{H}_{/(V/G)}(P_c \times_G V, X \times (V/G)) \end{aligned}

where again

$p \colon \mathbf{B}G \to \ast \,.$
###### Example

(symmetric powers)

Let

$G = \Sigma(n) \in Grp(Set) \hookrightarrow Grp(\infty Grpd) \overset{LConst}{\longrightarrow} \mathbf{H}$

be the symmetric group on $n$ elements, and

$V = \{1, \cdots, n\} \in Set \hookrightarrow \infty Grpd \overset{LConst}{\longrightarrow} \mathbf{H}$

the $n$-element set (h-set) equipped with the canonical $\Sigma(n)$-action. Then prop. says that right base change of any $p_\rho^\ast p^\ast X$ along

$\{1, \cdots, n\}/\Sigma(n) \longrightarrow \mathbf{B}\Sigma(n)$

is equivalently the $n$th symmetric power of $X$

$[\{1,\cdots, n\},X]/\Sigma(n) \simeq (X^n)/\Sigma(n) \,.$

## References

Standard textbook accounts include section A1.5.3 of

and section IV of

Last revised on February 25, 2020 at 14:51:53. See the history of this page for a list of all contributions to it.