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According to the nlab, the categorical trace of a 1-endomorphism $F:C\to C$ in a 2-category is the set hom$(1_C, F)$ of global elements of $F$. If $F$ is a functor in the 2-category Cat, the categorical trace is a set of natural transformations that assign to each object of $C$ a coalgebra of $F$ such that the obvious square commutes.

Any functor can be considered a special kind of profunctor; given an endofunctor, we can compute the coend of the corresponding profunctor.

Both of these concepts are generalizations of the trace, which for a function counts the number of fixpoints. In what sense do these "count" the fixpoints of a functor? I don't see how the categorical trace of a functor relates to fixpoints at all.

Also, does the notion of what constitutes a fixpoint change? The coend, in particular, seems like it might count an object $c$ as a fixpoint of $F$ if it's in the same endomorphism class rather than the same isomorphism class as $Fc$.

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What's a point? – Tom Goodwillie Jun 22 '11 at 2:19
omg, I first read "counit" instead of "count". – Martin Brandenburg Jun 22 '11 at 2:24
Without totally understanding your question, I am going to suggest looking at the very interesting paper of Ganter and Kapranov, which is certainly about categorical traces and about fixed points -- whether they are about YOUR categorical traces and YOUR fixed points I cannot say. – JSE Jun 22 '11 at 3:36
If this definition is compatible with the usual definition for monoidal categories with duals, then a nice example is the Lefschetz number of an endomorphism of a simplicial complex (regarded as a chain complex); the relationship with fixed points is given by the Lefschetz fixed point theorem. – Qiaochu Yuan Jun 22 '11 at 23:00
up vote 1 down vote accepted

Simon Willerton explains it all very well here:

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Here's a partial answer: in the case of an endofunctor $F$ on a discrete category $C$ (i.e. $F$ is a function), the coend of $F$ gives the set of fixpoints rather than the number: A profunctor $F:C \not\to C$ adds extra morphisms to $C$ so that the result is still a category. I'll say these morphisms are "in $F$". The coend of $F$ is the set of endomorphisms in $F$ mod conjugation by the morphisms in $C$; since the morphisms of $C$ are all identities, we just get the set of endomorphisms in $F$, i.e. fixed points of $F$.

The categorical trace doesn't reduce to anything useful in the case of a discrete category. A natural transformation $\alpha:1_C \Rightarrow F$ chooses for each $c \in C$ a morphism $\alpha_c:c \to Fc$ in $C$. Since we're assuming all the morphisms in $C$ are identities, $\alpha_c$ can't exist unless $Fc = c$. So it looks to me like the set hom$(1_C, F)$ is empty unless $F$ is the identity functor on $C$, in which case it's the terminal set.

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