If $f : \Bbb{Z}^{\times} \to (\Bbb{Z}/n)^{\times}$ is a monoid hom, then is $\Bbb{Z}^{\times} / \ker f \approx \operatorname{im} f$? For example $f(x) = x^2$.
That is can we quotient a monoid by the kernel of a monoid hom?
$\Bbb{Z}^{\times}$ the monoid not the group of units!
Let $f\colon M\to N$ be a monoid homomorphism. If by $\text{ker}(f)$, you mean $\{x\in M\mid f(x) = e\}$, where $e$ is the identity element of $N$, then $\text{ker}(f)$ is a submonoid of $M$, but it's not clear what $M/\text{ker}(f)$ means. There's not a general "quotient by a subobject" construction for monoids.
On the other hand, if by $\text{ker}(f)$ you mean the equivalence relation $\{(x,y)\mid f(x) = f(y)\}$, then $\text{ker}(f)$ is a congruence on $M$, and the quotient $M/\text{ker}(f)$ makes sense and is a monoid isomorphic to $\text{im}(f)$. This follows from the first isomorphism theorem of universal algebra (see here, and here for the special case of monoids).