Let
$$\begin{matrix} 0 & \rightarrow & A& \xrightarrow{f} & B & \xrightarrow{g} & C \\ & & \downarrow \alpha & &\downarrow \beta & &\downarrow \gamma\\ 0 & \rightarrow & A'& \xrightarrow{f'} & B' & \xrightarrow{g'} & C' \end{matrix}$$
be a commutative diagram of $R$-modules with exact rows. If $\beta, \gamma$ are isomorphisms, does it follow that $\alpha$ is surjective? I tried to prove this for awhile, but failed.
I was interested specifically in the following application: let $F \subseteq k$ be fields, $V(F), W(F)$ $F$-vector spaces, $V = k \otimes_F V(F),$ and $W = k \otimes_F W(F)$.
Regard $V(F), W(F)$ as $F$-subspaces of $V, W$. Let $f: V \rightarrow W$ be a linear transformation of $k$-vector spaces which maps $V(F)$ into $W(F)$. I'm trying to show that $\textrm{Ker } f$ is spanned by $\textrm{Ker } f \cap V(F)$ over $k$. My idea for the proof was to let $g$ be the restriction of $f$ to $V(W)$. This gives us an exact sequence
$$ 0 \rightarrow \textrm{Ker } f \cap V(F) \rightarrow V(F) \rightarrow W(F)$$
which when tensored with $k$ remains exact and fits into a commutative diagram
$$\begin{matrix} 0 & \rightarrow & k \otimes_F [\textrm{Ker } f \cap V(F)] & \xrightarrow{} & k \otimes_F V(F) & \xrightarrow{} & k \otimes_F W(F) \\ & & \downarrow & &\downarrow & &\downarrow \\ 0 & \rightarrow & \textrm{Ker } f & \xrightarrow{} & V & \xrightarrow{} & W \end{matrix}$$
By hypothesis the middle and right vertical arrows are isomorphisms, and the left arrow ought to be an isomorphism as well. Injectivity is clear, surjectivity seems to be more difficult.
This is standard diagram chasing: for $x \in A'$, show that $y = (f^{-1} \circ \beta^{-1} \circ f')(x)$ exists and has $\alpha(y) = x$.