If $g$ and $fg$ are isomorphisms, then so is $f$.

Question

Let's say you have a category, and you are looking at two arrows, $f$ and $g$, such that $fg$ is defined. You know that $g$ and $fg$ are isomorphisms, meaning there are arrows that are both their right and left inverses. What is the proof that $f$ is isomorphic? It appears that $g(fg)^{-1}$ would be the inverse, since

$$f(g(fg)^{-1})=(fg)(fg)^{-1}=Id_{IrrelevantPoint}$$

But I have no idea how to show that

$$(g(fg)^{-1})f=Id_{AnotherIrrelevantPoint}$$

If I could "distribute" the inverse, that would be helpful. I almost feel like what I am trying to proof is false, and the book is lying to me.

Note: In my book's definition at least, isomorphisms do not need to be unique. Take $(fg)^{-1}$ above to be an arbitrary inverse.

Note: In this section of the book, it was discussing monomorphisms and epimorphisms. I don't think it is relevant though.

Answer 1

Using the associative law to omit a lot of parentheses, and using that $gg^{-1}$ and $(fg)^{-1}fg$ are identity maps, which I abbreviate as $I$, we have $$g(fg)^{-1}f=g(fg)^{-1}fgg^{-1}=gIg^{-1}=I.$$

Answer 2

If $fg$ is an isomorphism and $g$ is an epimorphism, then $f$ is an isomorphism with inverse $h := g(fg)^{-1}$.

In fact, $fh=1$ is clear, and for $hf=1$ it suffices to check (since $g$ is epi) $hfg=g$, which is clear.

Note: It follows from the assumptions that $g$ actually is an isomorphism with inverse $(fg)^{-1} f$, but we don't need this for the proof.