0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '16 at 0:13
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.