1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $\ker(f_1)\to\ker(g_1)$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.