Do we still have a category of sets if the inverse image has more than one element? I'm kinda lost at one example in Awodey's: Category Theory.




I am trying to check this example, for that, I made a simple example function:

With this, I mean that there is one function $f$ which associates each element of the codomain. In this case, I can't have an inverse function - I'm not sure if this invalidates it being a category, but I faintly believe it does. Because it seems that in this universe of discourse, the arrows are functions. But I guess I could counter this by creating two pseudo-inverse functions (?!) in the following way:

It doesn't seem to violate the rules, except a little doubt for one of them: The definitions of composition. If I compose $f_{1}^{-1} \circ f$, I have:
$$cod(f)=dom(f_{1}^{-1})$$
$$cod(f_{1}^{-1})=dom(f)$$
I believe this is valid because he speaks about codomain instead of image. But it feels weird, $f_{1}^{-1}$ doesn't goes back entirely to $dom(f)$, that is: It doesn't takes all the elements of $dom(f)$.
Or perhaps I'm utterly confused/lost/stupid/in_need_of_help_for_mental_illness and completely missed the point.

The main problem is that the composition of functions with at most two preimages is not necessarily a function with at most two preimages.
In this wanna-be category, there does exist an arrow $\{1,2,3,4\}\to\{1,2\}$ (in fact several) and an arrow $\{1,2\}\to\{1\}$, but no arrow from $\{1,2,3,4\}$ to $\{1\}$.
For the restriction with finite pre-images, this problem does not occur, and you can verify that we indeed obtain a category.
