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Is there a usual way to express the concept of fixed points in category theory? What would I say to express that a morphism has a fixed point? Thank you!

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up vote 7 down vote accepted

A fixed point is a solution to the equation $x = f(x)$. The usual way to encode equations is as equalizers.

Given an object $A$ and an endomorphism $f : A \to A$, we can define "the subobject $B \mapsto A$ fixed by $f$" to be the equalizer of $f$ and the identity morphism. This subobject has the universal property that if $g : C \to A$ is any morphism with $fg = g$, then $g$ factors uniquely through $B$. (i.e. we can factor $g$ as $C \to B \to A$)

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Thank you for the prompt answer! I look forward to understanding it. – Nick Thomas Jan 20 '13 at 7:41

It is more interesting to ask whether an endofunctor has a fixed point. One distinguishes between least fixed points and largest fixed points. For example, the unit intervall is the largest bipointed topological space with a "self-similarity" $I = I \vee I$, and the set of natural numbers is the least set with $\mathbb{N} = 1 + \mathbb{N}$, the set of binary trees is the least set with $T=1+T^2$. For details, references and more examples, see:

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Is there some kind of lattice structure one exploits there? This looks reminiscent of the Tarski fixed point theorem. – Michael Greinecker Jan 27 '13 at 12:22
Actually Tarski's fixed point theorem is a special case of the Lemma (see the nlab) which says that an endofunctor $F$ on a category with $\omega$-chains which $F$ preserves has an initial algebra, applying this to the case of partial orders. – Martin Brandenburg Jan 27 '13 at 12:25
Thank you!${}{}{}{}{}{}{}{}$ – Michael Greinecker Jan 27 '13 at 12:29
By any chance, are the articles for the initial algebra and terminal coalgebra duals of each either? – PyRulez Feb 3 '15 at 10:56

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