# Extending monics in a commutative diagram

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?

• Always look at examples before diving into general category theory ... the example in the answer below is so easy that you would have seen this directly, too. Feb 25, 2014 at 20:11

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.
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.