Example of a homomorphism with a right or left inverse function that its right or left inverse is not a homomorphism

Question

If $f: A \to B$ is an injective or surjective homomorphism, then $f$ has a left or right inverse map respectively. The question is whether the right or left inverse function of $f$ is itself a homomorphism. I'm looking for examples that this doesn't happen.

Answer 1

This is pretty easy, let's just do it for some $\Bbb Z$-modules, i.e. abelian groups.

Let $A=\Bbb Z/2$ and $B=V=\Bbb Z/2\oplus \Bbb Z/2$ be the Klein 4-group.

Let $f(1)=(1,1)$. Then a left-inverse for $f$ is given by

$$\begin{cases}f_L:B\to A \\ (0,0),(1,0),(0, 1)\mapsto 0 \\ (1,1)\mapsto 1\end{cases}$$

clearly this cannot be a group homomorphism since we send a generating set to $0$, but the map is not the trivial map. It is also clearly a left-inverse for $f$.

For a surjective morphism, $f:A\to B$ we seek a right-inverse which is not a morhpism. This is really easy, though. We now let $A=V$ and $B=\Bbb Z/2$. Let

$$\begin{cases}f: A\to B \\ (1,1), (1,0)\mapsto 1 \\ (0,0),(0,1)\mapsto 0\end{cases}$$

Then a right inverse is

$$\begin{cases}f_R: B\to A \\ 0\mapsto (0,1) \\ 1\mapsto (1,1)\mapsto 1\end{cases}$$

The compositum is the identity on $B$, verifying the condition, but clearly the chosen inverse is not a homomorphism.