Pullback and Kernel

Question

We consider everything in the category of groups. It is known that monomorphisms are stable under pullback; that is, if $$\begin{array} AA_1 & \stackrel{f_1}{\longrightarrow} & A_2 \\ \downarrow{h} & & \downarrow{h'} \\ B_1 & \stackrel{g_1}{\longrightarrow} & B_2 \end{array} $$ is a pullback, then $g_1$ being one-to-one implies that $f_1$ is also one-to-one. Now if we weaken the condition, suppose that the kernel of $g_1$ is known, what can we say about the kernel of $f_1$? More precisely, if there is a commutative diagram

$$\begin{array} A & & B_0 & & A_1 &\stackrel{f_1}{\longrightarrow} & A_2\\ & & \parallel & &\downarrow{h}& &\downarrow{h'}\\ 0 & \stackrel{}{\longrightarrow} &B_0 & \stackrel{g_0}{\longrightarrow} &B_1 & \stackrel{g_1}{\longrightarrow} & B_2 & \stackrel{}{\longrightarrow} & 0 \end{array}$$

where the last row is an exact sequence and $A_1$ is the pullback, can we complete an exact sequence in the first row? Or at least is there a natural map from $B_0$ to $A_1$ making the diagram commutative?

Answer 1

In a pullback diagram we always have $\ker(f_1)\cong\ker(g_1)$.

For, we have the inclusion $\ker(g_1)\to B_1$ and the trivial map $\ker(g_1)\to A_2$. By the universal property of the pullback, we obtain a unique map $\ker(g_1)\to A_1$, whose image is then contained in $\ker(f_1)$. This gives the required inverse to the induced map $\ker(f_1)\to\ker(g_1)$.

Answer 2

There is an exact sequence $0 \to \ker(f_1) \cap \ker(h) \to \ker(f_1) \to \ker(g_1) \to 0$. I doubt that more can be said.