$$\begin{matrix} 0 & \rightarrow & A& \xrightarrow{f} & B & \xrightarrow{g} & C \\ & & \downarrow \alpha & &\downarrow \beta & &\downarrow \gamma\\ 0 & \rightarrow & A'& \xrightarrow{f'} & B' & \xrightarrow{g'} & C' \end{matrix}$$

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.

  • $\begingroup$ Oh..as usual, I figure out it's obvious right after I post the question. Assuming that $\alpha$ is injective, let $a' \in A'$, so $0 = \gamma^{-1} g' f'(a') = g \beta^{-1} f'(a')$, so $\beta^{-1} f'(a') = f(a)$ for some $a \in A$, so then $$f'(a') = \beta f(a) = f' \alpha(a)$$ which implies $a' = \alpha(a)$ $\endgroup$
    – D_S
    Sep 10, 2016 at 0:13

1 Answer 1


This is standard diagram chasing: for $x \in A'$, show that $y = (f^{-1} \circ \beta^{-1} \circ f')(x)$ exists and has $\alpha(y) = x$.


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