Mastering proof by contradiction, no morphism can have more than one inverse.

Question

Problem:

1. I am a total beginner to category theory, so might abuse common notation or terminology at some point without even being aware of that. So please stay sharp. My primary goal is to master notation, make myself clear and understandable by others.
2. I know my proof technique is insufficient, I haven't had enough practice; so I'll try to perform a fully correct proof by contradiction and would like to get some feedback whether it looks like a real proof or not. If not, it would be really great to see how it supposes to look.

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Show that a map in a category can have at most one inverse.

Given somewhat category $\mathbb{C}$ with $A, B \in \mathsf{Ob}(\mathbb{C}), A \not = B$ and $f \in \mathsf{Hom}_{\mathbb{C}}(A, B)$, there can't be more than a single $g \in \mathsf{Hom}_{\mathbb{C}}(B, A)$ such that $g \circ f = 1_A, f \circ g = 1_B$.

Since I decided to prove the statement above by contradiction, I start from assumption that there indeed exists $g' \in \mathsf{Hom}_{\mathbb{C}}(B, A), g' \not = g$ such that $g' \circ f = 1_A, f \circ g' = 1_B$. Actually, I kind of uncertain here, because there is no explicit way how to calculate equality of morphisms; maybe there is one, however I did not mention it and thus $g' \not = g$ seems to be useless.

Nevertheless: $$g \circ f = 1_A = g' \circ f$$ $$\Rightarrow g \circ f = g' \circ f$$ $$\Rightarrow g = g'$$

Hence, $g = g'$ and $g \not = g'$ is contradiction. Thus, my initial assumption can't be true; thus, $g = g'$; thus, there indeed can't be more than a single inverse of $f$.

Answer 1

If $g\circ f=g'\circ f$ then the conclusion $g=g'$ that you make is only justified under the extra condition that $f$ is right-cancellable, or - in terms of categories - that $f$ is an epimorphism.

Correct route:

If $g\circ f=1_A$ and $f\circ g=1_B$ and next to that also $g'\circ f=1_A$ then$$g'=g'\circ1_B=g'\circ (f\circ g)=(g'\circ f)\circ g=1_A\circ g=g$$

Answer 2

I don't yet have the reputation to write a comment, so I'll just write that here: You don't need the assumption, that the objects $A$ and $B$ are not equal. It is no problem to have morphisms $A \to A$ in a category.

Now it is not in general true, that $g \circ f = g' \circ f \Rightarrow g = g'$. Therefore you should not use such an argument.

If you want a little hint: Try to start with $g = ...$ and deduce that $g = g'$ just by the information you've got.

//Edit: Not really important, but maybe still worth mentioning: The symbol $\mathbb{C}$ is very non-standard for a category.