Isomorphism: minimal number of steps to check There are in general two methods to check a map is a isomorphism.
The first is to show it is injective, surjective and a homomorphism. A second way is via the construction of inverse.
I got kind of confussed about this second method because some people check more things than others. What are the minimal steps one has to make to follow this second method?
My guess is:

*

*show $f$ is bijective

*find $g$ such that $fg=id$
Are there any steps missing? That is, does it depend on the source some people are also checking things like surjectivity and injectivity of $g$? Is this necessary?

The "second way" is insufficient.
For example, a group isomorphism must map the identity in its domain to the identity in its codomain.
Proof: Let $x\in G$ and $\varphi: G\to H$ be a group isomorphism. Then
$$\begin{align} 
\varphi(x)&=\varphi(e_Gx)\\
&=\varphi(e_G)\varphi(x),
\end{align}$$
so if we multiply on the right by $(\varphi(x))^{-1}$, we get
$$\begin{align}
e_H&=\varphi(x)(\varphi(x))^{-1}\\
&=\varphi(e_G)(\varphi(x)(\varphi(x))^{-1})\\
&=\varphi(e_G)e_H\\
&=\varphi(e_G).
\end{align}$$
Since $\varphi$ is an isomorphism, in particular, it is a bijection, so $e_G$ is the only $g\in G$ for which $\varphi(g)=e_H$.$\square$
But consider
$$\begin{align}
f: \Bbb Z_2&\to \Bbb Z_2,\\
[0]_2&\mapsto [1]_2,\\
[1]_2&\mapsto [0]_2.
\end{align}$$
Clearly $f$ has an inverse (namely, itself), but does not map the identity to the identity, so cannot be an isomorphism.
