On page 7 of Topology: A Categorical Approach by Bradley, Bryson and Terilla I am asked to verify what I've written in the title of this post.
My attempt:
Reflexivity:
The isomorphism from $X$ to $X$ is $\text{id}_X$, since $\text{id}_X$ is its own inverse. Therefore $X \cong X$.
Symmetry:
If $X \cong Y$ that means there is an invertible morphism $f \colon X \to Y$. Say its inverse is $g \colon Y \to X$. When we show that $g$ is the left and right inverse of $f$, we are showing that $f$ is the right and left inverse of $g$ (respectively), so that $f$ is the inverse morphism of $g$ and therefore $Y \cong X$.
Transitivity:
Suppose we have objects $X, Y$ and $Z$ such that $X \cong Y$ and $Y \cong Z$. There are thus invertible morphisms $f \colon X \to Y$ and $g \colon Y \to Z$. By the composition axiom of categories, we know that $gf \colon X \to Z$ is a morphism. We want to show that $gf$ is invertible, hence $X \cong Z$.
We can compose the inverse morphisms $f^{-1} \colon Y \to X$ and $g^{-1} \colon Z \to Y$ to get $f^{-1}g^{-1} \colon Z \to X$. Composing $f^{-1}g^{-1}$ with $gf$, we get
\begin{align*} (f^{-1}g^{-1})(gf) &= f^{-1}(g^{-1}g)f\\ &= f^{-1}\text{id}_Y f\\ &= f^{-1}f\\ &= \text{id}_X, \end{align*}
and similarly on the right side, hence $X \cong Z$ as desired.
Question:
Do I need to prove some kind of generalized associativity to justify the manipulations I used to prove transitivity? The category axiom for associativity only says that $f(gh) = (fg)h$ for suitably defined morphisms.
I appreciate any help.