Given a commutative diagram of vector spaces with exact rows
I'd like to understand why knowing that $h$ is injective proves that the injection $ coker (\alpha) \to coker (\beta)$ is equivalent to $h(A) \cap \beta (V) = \beta f (U)$. Thanks a lot for any help.
A typical element of the cokernel of $\alpha$ is the coset $a+\alpha(U)$. A typical element of the cokernel of $\beta$ is the coset $b+\beta(V)$.
The map $h^*$ from the cokernel of $A$ to that of $B$ induced by $h$, is injective if and only if $h(a)+\beta(V)=\beta(V)\implies a+\alpha(U)=\alpha(U)$, that is if and only if $h(a)\in\beta(V)\implies a\in\alpha(U)$.
Suppose $h^*$ is injective and $b\in h(A)\cap\beta(V)$. Then $b=h(a)$ for some $a\in A$ and so $a\in\alpha(U)$. That is $a=\alpha(u)$ for $u\in U$ and $b=h(\alpha(u))=\beta f(\beta(u))$ so $b\in f\beta(U)$. Thus $h(A)\cap\beta(V)\subseteq f\beta(U)$. The reverse inclusion is easy.
Conversely suppose $h(A)\cap\beta(V)=f\beta(U)$. If $h+\alpha(U)$ is in the kernel of $h^*$, then $h(a)=\beta(v)$ for $v\in V$. Then $h(a)\in h(A)\cap\beta(V)=\beta f(U)=h\alpha (V)$. So $h(a)=h(a')$ with $a'\in \alpha(V)$ and as $h$ is injective $a=a'\in\alpha(U)$ so $a+\alpha(U)=\alpha(U)$ is zero in the cokernel. Therefore $h^*$ is injective.
It's really just a diagram chase.