Exercise 0.3 (b) in Topology: A Categorical Approach by Bradley, Bryson and Terilla asks for an example of an epimorphism that is not right-invertible. Before I get to that, I am trying to understand the example given of a monomorphism that is not left-invertible. Below is a quote from page 14:
For example, the map $n \mapsto 2n$ defines a left-cancellative group homomorphism $f \colon \mathbb Z / 2 \mathbb Z \to \mathbb Z / 4 \mathbb Z$. However, there is no group homomorphism $g \colon \mathbb Z / 4 \mathbb Z \to \mathbb Z / 2 \mathbb Z$ so that $gf = \text{id}_{\mathbb Z / 2 \mathbb Z}$.
I try to explain this to myself as follows. We have a homomorphism $f$ such that $f([0]_2) = [0]_4$ and $f([1]_2) = [2]_4$. The text says on page 14 that a function is injective $\Leftrightarrow$ function is left-cancellative. Therefore, the morphism (function) $f$, clearly being injective, must be left-cancellative.
Unfortunately I cannot justify the assertion that there is no $g$ such that $gf = \text{id}_{\mathbb Z / 2 \mathbb Z}$. Perhaps the most succinct reason why I'm struggling with the idea that $f$ is not left-invertible is that group homomorphisms are in fact functions, and the text says that a function is injective $\Leftrightarrow$ it has a left inverse.
I figure if I can't understand this example, I have no hope of giving an example of an epimorphism that is not right-invertible. I appreciate any help.
Edit: epimorphism that is not right-invertible
I would appreciate verification that this example is correct. Consider the group homomorphism $h \colon \mathbb Z \to \mathbb Z / 3 \mathbb Z$ given by $z \mapsto z \mod 3$. Because this morphism (function) is surjective, it is right-cancellative, i.e. epic.
We want to show that there is no group homomorphism $g \colon \mathbb Z / 3 \mathbb Z \to \mathbb Z$ such that $hg = \text{id}_{\mathbb Z / 3 \mathbb Z}$. I was able to show in a few lines that $g$ must be the trivial homomorphism. Therefore, for example, $h \big( g ( [2]_3 ) \big) = h(0) = [0]_3 \neq [2]_3$, as desired.
Best Answer
Any such $g$ must take $[1]_4$ to $[0]_2$ or $[1]_2$. It therefore takes $[2]_4$ to $[0]_2$. But $f([1]_2)=[2]_4$, so $(g\circ f)([1]_2)=g([2]_4)=[0]_2\ne[1]_2$.