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 $$ \operatorname{Img}(\psi') = \mathbb{Z}\oplus 0 = \ker(\phi'). $$