Show that if $f$,$g$ are iso arrows, then $f\circ g$ is too. Could anyone check my proof please?

Show that if $f$, $g$ are iso arrows, then $f\circ g$ is too, with $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$.

R Goldbatt's Topoi: $f:a\to b$ is an iso arrow (or invertible) if there is an arrow $g: b\to a$ such that $g\circ f=1_a$ ($1_a$ being the identity arrow on $a$) and $f\circ g=1_b$.
So let's assume that $g: a\to b$ and $f: b\to c$ (I know it's confusing, but the other way round seems even more so...), then the task is to find an arrow $h: c\to a$ such that $h\circ f\circ g=1_a$ and $f\circ g\circ h=1_c$.
It would seem that $f\circ g^{-1}$ is a good candiate for $h$. So I need to show that $(g^{-1}\circ f^{-1})\circ (f \circ g)=1_a$.
Let's beging with $(g^{-1}\circ f^{-1})\circ (f \circ g)(a)$. But because $f$ is iso, let $f^{-1}$ be that inverting arrow:
$(g^{-1}\circ 1_b) \circ g(a)$. But composing with identity just gives you the function, so we can take $1_b$ out here, i.e.
$g^{-1}\circ g(a)$; but since $g$ is also iso we apply the same trick so we end up with $1_a(a)$.
$f\circ g\circ h=1_c$ should be able to be proved in a similar manner.
 A: In general you got the right idea to use the inverses of $f$ and $g$ to create an inverse for $f \circ g$. The essence of the proof is also there, but there are a few issues with the presentation that would really make the proof harder to read for someone who doesn't know how it would go on beforehand.
First you say $f \circ g^{-1}$ is a good candidate, but I would read this as $f \circ (g^{-1})$, which would definitely not work and is also not what you use. It would not even compose, since $g^{-1}: b \to a$ and $f: b \to c$. If you mean $(f \circ g)^{-1}$, then you get into a circular argument. You are currently trying to show that the inverse of $f \circ g$, which we denote by $(f \circ g)^{-1}$, exists.
Luckily you continue working with $g^{-1} \circ f^{-1}$, which does exist by the assumption of $f$ and $g$ being isomorphisms. This will indeed be the inverse of $f \circ g$. Then you want to show that $g^{-1} \circ f^{-1} \circ f \circ g = 1_a$ (tip, you can omit the parentheses because composition is associative). The key idea is there, namely:
$$
g^{-1} \circ f^{-1} \circ f \circ g =
g^{-1} \circ 1_b \circ g =
g^{-1} \circ g =
1_a.
$$
However, you write "$(a)$" after all these things, as if you are applying a function to some element $a$. This is not correct, arrows in a category are not necessarily functions of sets. So you cannot apply them to elements. The way to reason is often very similar though. The thing I just wrote here is really the same as what you wrote, just without the "$(a)$" bits.
So long story short: good insights, the spirit of the proof is correct, just some issues with the presentation. I would say you are on the right track of learning this stuff!
