Sets are my objects and as arrows, those $ f : A \rightarrow B $ such that for all $ b \in B $, the subset $ f^{-1}(b) \subseteq A $ has at most two elements (rather than one). I want to prove whether it is a category.
NOTE : Please also points out deficiency in my mathematical language (argument).
This is my attempt.
Identity 'morphism' can be proven by showing that it exists. A simple 'one-to-one mapping' from a set to itself will do the job.
To form a concrete example to show one instance of 'composition', I take three sets $A = \{1, 2, 3 \} $, B = { p, q }, C = { a, b, c }. Let $ f : A \rightarrow B $ and $ g : B \rightarrow C $ then $ f = \{(1, w), (2, w), (2, x), (3, x) \}$ and $ g = \{ (w, a), (w, b), (x, a), (x ,c) \} $. Now I try to compose $f$ and $g$. Since both $f$ and $g$ multi-valued 'function', I am not sure whether following is the right step. $ f \bullet g = \{ (1,a), (1, b), (2, a), (2, b), (2, a), (2, c), (3, a), (3, c) \} = \{ (1,a), (1, b), (2, a), (2, b), (2, c), (3, a), (3, c) \} $. This indeed is some morphism $ X : A \rightarrow C $. But it violates the given constrains on co-domain on the 'function' since for $a \in C $, $ X^{-1}(a) $ has more than two elements. They are $\{1, 2, 3\}$.
Thus, this thing is not a category
Now I am having trouble with finite case, namely when $f^{-1}(b)$ is finite, or infinite?
I have constructed a proof for finite case on similar arguments and it proves that this is a category.
What about infinite case? Can I prove it without using the concept from vector spaces?