# Categories of endofunctions

Endofunctions on a set $X$ form a monoid $End(X)$, which is - as every monoid - a category with a single object $X$.

There seems (to me) to be another conceivable category $E$ of endofunctions on $X$:

Consider as objects of $E$ the elements of $X$. Consider as morphisms of $E$ the ordered pairs $(x,f)$ with $x \in X$ and $f: X \rightarrow X$.

• The source of $(x,f)$ is $x$, the target of $(x,f)$ is $fx$.

• Composition: $(x,f)\circ(y,g) = (y,f\circ g)$
(if defined, i.e. if $x = gy$)

• Identities: $\text{id}_x = (x,\text{id})$

Such an "endocategory" $E$ has - among others - the following characteristics;

• Each object has the same number of out-arrows ("$|X|^{|X|}$").

• There is an obvious equivalence relation between the morphisms: $(x,f) \simeq (x',f')$ iff $f = f'$.

Question 1: Can this equivalence relation be defined categorically ("by dots and arrows") for an endocategory $E$?

Question 2: Where and under which name do endocategories play a role? Are they of any interest?

Question 3: How can endocategories be characterized?

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I think this category isn't really interesting. All objects are isomorphic. There are lots of morphisms. And after all, we only input a set. Probably more interesting categories arise when we start with an object $X$ of an interesting category, choose some of the various meanings of "element" (mathoverflow.net/questions/98139) and see what happens. When $x,y$ are elements of $X$, then they are isomorphic in the endocategory if there is an automorphism $f : X \to X$ mapping $f(x)=y$. Thus we end up with orbits of $\mathrm{Aut}(X)$. – Martin Brandenburg Oct 22 '12 at 10:25
This is basically the category of elements of the canonical copresheaf on the monoid $\textrm{End}(X)$. – Zhen Lin Oct 22 '12 at 18:54
@Martin: Why is this category less interesting than End(X), or vice versa: why is End(X) more interesting? – Hans Stricker Oct 22 '12 at 20:24
Consider the category $Set_{\ast}$ of pointed sets (i.e. pairs $(X,x)$ where $x \in X$ and morphisms preserve the distinguished element). For any fixed set $X$, the subcategory of $Set_{\ast}$ where objects are pairs $(X,x)$ for all $x$ is effectively the endocategory of $X$. – sdcvvc Oct 23 '12 at 12:24

I can only answer your first question. As to that, the equivalence relation cannot be defined without reference to the labels on your arrows. Given one arrow $f_x: x\mapsto y,$ I'd like to match it to exactly one element of $\hom_E(z,w)$. Yet every such arrow combines with $f_x$ to define $|X|^{|X|-2}$ different functions, with $|X|-2=|X|$ if $|X|$ is infinite. That is, the classification problem isn't well-posed at all.