$h$ is a homomorphism, $f$ is any map and $a \neq b$. Then $h(f(a)) \neq h(f(b)) \implies f(a) \neq f(b)$

Question

Let $h$ is a homomorphism, $f$ is any map and $a \neq b$. Then $h(f(a)) \neq h(f(b)) \implies f(a) \neq f(b)$. This was a step in a proof when I was reading Free modules (to prepare for the so called universal mapping theory). In the setting, $f$ is a map between a set $S$ and R-module $F$ from Sze-Tsien Hu's book.

Answer 1

In general, $g(a) \neq g(b) \implies a \neq b$.

Answer 2

The contrapositive is plainly true:

$f(a) = f(b) \implies h(f(a)) = h(f(b))$