Prove that that the inverse to a map is unique.

Question

I'm reading Conceptual Mathematics: An Introduction to Categories by Schanuel and so far, it's only mentioned one method of showing that two maps $f$ and $g$ are equal and that is by showing that if for every point $a:1 \to A$, $f \circ a = g \circ a$, then $f = g$. Then I am asked to show that if two functions $g: B \to A$ and $k: B \to A$ are inverses to a function $f: A \to B$, then $g = k$.

So far, all I have is that if $g$ and $k$ are both inverses to $f$, then $f \circ g = f \circ k = \mathbf 1_A$ and $g \circ f = k \circ f = \mathbf 1_B$, however this doesn't seem to help very much. I'm wondering, can I say that because I know that Dom(g) = Dom(k) $\land$ Codomain(g) = Codomain (k) $\land$ $f \circ g = f \circ k$ that it must be the case that g = k?

Answer 1

The argument you propose does not hold in general. We say $f$ is a monomorphism if $f\circ g=f\circ k$ implies $g=k$ for any (valid) $g,k$. In the case where $f$ is an isomorphism, then $f$ is a monomorphism, which can be shown by the following:

Since $f\circ g=f\circ k$, we must have $$g\circ(f\circ g)=g\circ(f\circ k),$$ and by associativity this is $$\mathbf{1}_B\circ g=(g\circ f)\circ g=(g\circ f)\circ k=\mathbf{1}_B\circ k,$$ hence $g=k$..