Extending monics in a commutative diagram

Question

Given a commutative diagram in a Grothendieck category $\mathscr{A}$

\begin{array}{ccccccccc} 0 & \longrightarrow & A' & \overset{i}{\longrightarrow} & A & \overset{p}{\longrightarrow} & A'' & \longrightarrow & 0\\ & & f\downarrow & & g\downarrow & \\ 0 & \longrightarrow & B' & \overset{j}{\longrightarrow} & B & \overset{q}{\longrightarrow} & B'' & \longrightarrow & 0 \end{array}

with $f: A' \rightarrow B'$ and $g: A \rightarrow B$ , There exists a unique arrow $h: A'' \rightarrow B''$ making the large diagram commute.

My question is, if f and g are monic, does this force h to also be monic?
If so .. why? and if not would assuming g to be an isomorphism do the trick?

Answer 1

Even if $g$ is an isomorphism, $h$ need not be monic. Consider

$$\begin{array}{ccccccccc} 0 & \longrightarrow & 0 & \longrightarrow & \mathbb{Z} & \longrightarrow & \mathbb{Z} & \longrightarrow & 0\\ & & \downarrow & & \text{id}\downarrow & & h\downarrow \\ 0 & \longrightarrow & \mathbb{Z} & \longrightarrow & \mathbb{Z} & \longrightarrow & 0 & \longrightarrow & 0 \end{array}$$

However, if $g$ is monic and $f$ is epi, then $h$ is monic, and can be proved using a diagram chase. This is part of the 5-lemma.

Answer 2

No in general, for example: if g is an isomorphism then by the snake lemma, $Ker(h)\cong Coker(f)$ and $0\cong Coker(h)$, so h is monic, iff f is the zero arrow.