nLab
initial algebra

An initial algebra for an endofunctor $F$ on a category $C$ is an initial object in the category of algebras of $F$.

If $F$ has an initial algebra $\alpha :F(X)\to X$, then $X$ is isomorphic to $F(X)$ via $\alpha$; this is Lambek’s theorem. In this sense, $X$ is a fixed point of $F$. Being initial, $X$ is the smallest fixed point of $F$ in that there is a map from $X$ to any other fixed point (indeed, any other algebra), and this map is an injection if $C$ is Set.

The dual concept is terminal coalgebra, which is the largest fixed point of $F$.

Proof of Lambek’s theorem

Given an initial algebra structure $\alpha :F(X)\to X$, define an algebra structure on $F(X)$ to be $F(\alpha ):F(F(X))\to F(X)$. By initiality, there exists an $F$-algebra map $i:X\to F(X)$, so that

$$\begin{array}{ccc}F(X)& \stackrel{F(i)}{\to }& F(F(X))\\ \alpha \downarrow & & \downarrow F(\alpha )\\ X& \underset{i}{\to }& F(X)\end{array}$$\array{ F(X) & \overset{F(i)}{\to} & F(F(X)) \\ \alpha \downarrow & & \downarrow F(\alpha) \\ X & \underset{i}{\to} & F(X) }

commutes. Now it is trivial, in fact tautological that $\alpha$ is itself an $F$-algebra map $F(X)\to X$. Thus $\alpha \circ i=1_X$, since both sides of the equation are $F$-algebra maps $X\to X$ and $X$ is initial. As a result, $F(\alpha )\circ f(i)=1_{F(X)}$, so that $i\circ \alpha =1_{F(X)}$ according to the commutative square. Hence $\alpha$ is an isomorphism, with inverse $i$.

Examples

In many cases, initial algebras can be constructed in recursive fashion, using the following special case of a theorem due to Adámek.

Theorem (Adámek)

Let $C$ be a category with an initial object $0$ and colimits of chains $\omega \to C$ (where $\omega$ is the first infinite ordinal), and suppose $F:C\to C$ preserves colimits of $\omega$-chains. Then the colimit $\gamma$ of the chain

$$0\stackrel{i}{\to }F(0)\stackrel{F(i)}{\to }\dots \to F^{(n)}(0)\stackrel{F^{(n)}(i)}{\to }{F}^{(n+1)}(0)\to \dots$$0 \overset{i}{\to} F(0) \overset{F(i)}{\to} \ldots \to F^{(n)}(0) \overset{F^{(n)}(i)}{\to} F^{(n+1)}(0) \to \ldots

carries a structure of initial $F$-algebra.

The $F$-algebra structure $F\gamma \to \gamma$ is inverse to the canonical map $\gamma \to F\gamma$ out of the colimit (which is invertible by the hypothesis on $F$). The proof of initiality may be extracted by dualizing the corresponding proof given at terminal coalgebra.

This theorem applies in particular to any functor $F:\mathrm{Set}\to \mathrm{Set}$ which is a colimit of finitely representable functors $\mathrm{hom}(n,-):X\mapsto X^n$, as in the following examples.

• Let $A$ be a set, and let $F:\mathrm{Set}\to \mathrm{Set}$ be the functor $F(X)=1+A\times X$. Then the initial $F$-algebra is $A^{*}$, the free monoid on $A$. The $F$-algebra structure is

  $$(e,m\mid ):1+A\times A^{*}\to A^{*}$$(e, m| ): 1 + A \times A^* \to A^*

  where $e:1\to A^{*}$ is the identity and $m\mid :A\times A^{*}\to A^{*}$ is the restriction of the monoid multiplication along the evident inclusion $i\times 1:A\times A^{*}\to A^{*}\times A^{*}$.

  This “fixed point” of $F$ can be thought of as the result of the (slightly nonsensical) calculation

  $$1+A\times X=X\Rightarrow X=\frac{1}{1-A}=1+A+A^2+\dots =A^{*}$$1 + A \times X = X \Rightarrow X = \frac1{1 - A} = 1 + A + A^2 + \ldots = A^*

  which can be made rigorous by interpreting the initial equality as defining the solution $X$ by recursion, and applying the theorem above.

• Let $F:\mathrm{Set}\to \mathrm{Set}$ be the functor $F(X)=1+X^2$. Then the initial $F$-algebra is the set $T$ of isomorphism classes of finite (planar, rooted) binary trees. The $F$-algebra structure is

  $$(r,j):1+T^2\to T$$(r, j): 1 + T^2 \to T

  where $r:1\to T$ names the tree consisting of just a root vertex, and $j:T^2\to T$ creates a tree $t\vee t\prime$ from two trees $t$, $t\prime$ by joining their roots to a new root, so that the root of $t$ becomes the left child and the root of $t\prime$ the right child of the new root.

  The recursive equation

  $$T=1+T^2$$T = 1 + T^2

  would seem to imply that the structure $T$ behaves something like a structural “sixth root of unity”, and indeed the structural isomorphism $T\cong F(T)$ allows one to exhibit an isomorphism

  $$T=T^7$$T = T^7

  constructively, as famously explored in the paper by Andreas Blass, Seven Trees in One.

• Let $F:\mathrm{Set}\to \mathrm{Set}$ be the functor $F(X)=X^{*}$ (the free monoid from an earlier example). Then the initial $F$-algebra is the set of isomorphism classes of finite planar rooted trees (not necessarily binary as in the previous example).

• Let $C$ be the coslice category $\mathbb{Z}\downarrow \mathrm{Ab}$, and let $F:C\to C$ be the functor which pushes out along the multiplication-by-$p$ map $p\cdot -:\mathbb{Z}\to \mathbb{Z}$. Then the initial $F$-algebra is the Pruefer group $\mathbb{Z}[p^{-1}]/\mathbb{Z}$. See the discussion at the n-Category Café, starting here.

• Let $\mathrm{Ban}$ be the category of complex Banach spaces and maps of norm bounded above by $1$, and let $F:ℂ\downarrow \mathrm{Ban}\to ℂ\downarrow \mathrm{Ban}$ be the squaring functor $X\mapsto X\times X$. Then the initial $F$-algebra is $L^1([0,1])$ (with respect to the usual Lebesgue measure). This result is due to Tom Leinster; see this MathOverflow discussion.