Every identity arrow is iso

Question

Every identity arrow is iso.

I struggling a bit with this proof, can someone help?

I understand that isomorphic is invertible and for a function $f:a\to b$ is invertible if there exists a function $g:b\to a$ s.t. $g\circ f=1_a$ and $f\circ g=1_b$.

Answer 1

Consider an identity arrow $1_x$ for some object $x$. By definition, $1_x: x\to x$ and for every $f:y\to x$ and every $g:x\to z$, we have $1_x\circ f=f$ and $g\circ 1_x=g$; in particular, then, with $f=g=1_x$, we have $1_x\circ 1_x=1_x$. Can you continue from here?

Hint:

Prove uniqueness of inverses.