# Showing that a certain map in a commutative diagram with exact rows is injective

exact-sequence

This exercise is from Dummit and Foote, Section 10.5. (Exercise 1, part d)

The following diagram is commutative with exact rows. We know that $$\alpha,\gamma$$ are surjective, and $$\beta$$ is injective. I want to show that $$\gamma$$ is injective.

I tried starting with some $$c \in C$$ with $$\gamma c = 0$$. But then I get stuck in $$C'$$. Since I don't know anything about the injectivity/surjectivity of $$\phi$$ and $$\phi'$$, I don't know how to get out of $$C'$$ and use exactness or commutativity.

First I assumed that there exists some $$b \in B$$ such that $$\phi(b) = c$$. Then by commutativity of the right square we have
$$\gamma\phi b = \phi'\beta b = 0$$
Therefore $$\beta b \in \ker \phi' = \mathrm{Im(\psi')}$$ so we have $$\psi' a' = \beta b$$ for some $$a' \in A'$$. We also know that $$\alpha$$ is onto so we have $$\alpha a = a'$$ for some $$a \in A$$. Now we will use the commutativity of the left square:
$$\psi' \alpha a = \beta \psi a = \beta b$$
Therefore $$\beta \psi a = \beta b$$ and by $$\beta$$'s injectivity we have $$\psi a = b$$. Now we apply $$\phi$$:
$$\phi \psi a = 0 = \phi b = c$$
I used exactness in the last part.

But I don't know what to do when $$c$$ is not in $$\phi$$'s image. Hints would be appreciated. Thanks in advance.

I believe the statement is false: Let $$A=B=C=B'=\mathbb{Z}\oplus \mathbb{Z}$$ and $$A'=C'=\mathbb{Z}$$. Consider the following maps: $$\psi(x,y)=(x,0), \quad \phi(x,y)=(0,y)$$ $$\alpha(x,y)=x, \quad \beta(x,y)=(x,y), \quad \gamma(x,y)=y$$ $$\psi'(x) = (x,0), \quad \phi'(x,y)=y.$$
Then it is easy to see that the diagram commutes, that $$\alpha$$ and $$\gamma$$ are surjective and $$\beta$$ is injective but $$\gamma$$ is not injective. Also, the rows are exact because $$\operatorname{Img}(\psi) = \mathbb{Z}\oplus 0 = \ker(\phi)$$ and $$\operatorname{Img}(\psi') = \mathbb{Z}\oplus 0 = \ker(\phi').$$