# The kernel of a homomorphism at the end of an exact sequence

Suppose that $$A,B,$$ and $$C$$ are groups and there is an exact sequence $$A \to B \to C \to 1$$, and $$A_1,B_1,$$ and $$C_1$$ are also groups and there is an exact sequence $$A_1 \to B_1 \to C_1 \to 1$$. So $$B \to C$$ and $$B_1 \to C_1$$ are surjective but $$A \to B$$ and $$A_1 \to B_1$$ are not necessarily injective. Suppose there are also homomorphisms $$\alpha \colon A \to A_1$$ and $$\beta \colon B \to B_1$$ and $$\gamma \colon C \to C_1$$ such that the resulting diagram commutes. If the kernels of $$\alpha$$ and $$\beta$$ are finite, must the kernel of $$\gamma$$ be finite?

I don't see why it would be true, but I don't know any counterexample. I thought about it for a while and can't come up with any simple proof. My guess is that it's not true.

$$\newcommand\twoheaduparrow{\mathrel{\rotatebox{90}{\twoheadrightarrow}}} \begin{array} A & \mathbb{Z}_2 & {\hookrightarrow} & \mathbb{Z}_2 &\oplus& \mathbb{Z} &{\twoheadrightarrow} & \mathbb{Z} & \ \\ & \downarrow{} & & &\downarrow{}& &&\downarrow{}\\ & 2\mathbb{Z} & \stackrel{}{\hookrightarrow} && \mathbb{Z} & \stackrel{}{\twoheadrightarrow} &&\mathbb{Z}_2 & & \end{array}$$
Here $$\ker{\gamma} = 2\mathbb{Z}$$, so, infinite. Note the first downward map is just the $$0$$ map which commutes in that square since on the right side of the square we embed then project, so $$\mathbb{Z}_2$$ goes to $$0$$ as well.