Six term exact sequence

Question

I'm trying to understand the proof of this corollary $5.4$, but the proof itself is too short. It's based on theorem $5.1$, which is:

I'm trying to relate the $g$ from the corollary to the theorem. In the theorem, when $g$ is trivial, $f$ and $h$ are epimorphism and monomorphism. If at least $g$ were trivial, I'd have that it's at least an homomorphism.

I'm trying to look the other way: supposing $f$ and $h$ are trivial, but then they're not mono neither epi, so this wouldn't help either...

For $b\iff c$, I think that $f$ and $h$ being trivial, then $\ker k = \{0\} \implies k$ mono. Now $f$ being trivial, $\ker f = im d= 0$, right? If the image is $0$, how can it be epi? Something's wrong here.

Answer 1

Let's focus on this "portion" of the sequence: $$A\overset d \longrightarrow B\overset f\longrightarrow C\overset g\longrightarrow D$$

We apply theorem 5.1 to $f$. By this theorem $f$ is the trivial homomorphism iff $d$ is an epimorphism iff $g$ is a monomorphism.

Analogously, if we consider the "portion": $$C\overset g\longrightarrow D\overset h\longrightarrow E\overset k\longrightarrow F$$

Applying theorem 5.1 to $h$ we find that $h$ is the trivial homomorphism iff $k$ is a monomorphism iff $g$ is an epimorphism.

From the above follows immediately that (a)$\iff$(b) and (b)$\iff$(c).

Answer 2

If $f$ is trivial then $ker(f) = B$, but from the very definiotion of exact sequence you know that $ker(f) = im(d)$ hence $d $ is epi. I think you may have confused the notion of an image and kernel.

$f$ is trivial which means that it sends everything to zero - hence every element of the domain lies in the kernel