Images in a short exact sequence Suppose 
$$
0\to V\to W\to X\to 0\\
\downarrow\quad\quad\downarrow\quad\quad\downarrow\\
0\to V'\to W'\to X'\to 0\\
$$
is a commutative diagram of vector spaces, with the top and bottom rows short exact sequences.  When is it true that I have an exact sequence
$$
0\to im(V\to V')\to im(W\to W')\to im(X\to X')\to 0?
$$
I think I get from the snake lemma that this occurs if and only if the connecting morphism from the snake lemma is the trivial map, but I'm not feeling particularly confident in my answer, or maybe there's something simpler.
 A: Let us denote the vertical maps by $f,g,h$, and the horizontal ones by $i,p$ resp. $i',p'$. Then one checks directly that $im(f) \to im(g)$ is mono (since $i'$ is) and that $im(g) \to im(h)$ is epi (since $p$ is). Besides, it is clear that we have a complex. Thus, the sequence is exact iff the kernel of $im(g) \to im(h)$ is contained in the image of $im(f) \to im(g)$. The kernel equals $im(g) \cap im(i')$, and the image equals $im(gi)=im(i'f)$. A sufficient condition for equality is that $f$ is epi (clear), but $h$ mono also suffices: If $w' \in im(g) \cap im(i')$, write $w' =  g(w)$, then $0=p'(w')=h(p(w))$, hence $0=p(w)$, i.e. $w \in im(i)$ and $w' \in im(gi)$.
Edit. Your guess is correct. Recall that the connecting homomorphism $\delta : ker(h) \to coker(f)$ is defined as follows: If $x \in ker(h)$, write $x=p(w)$, then $p'(g(w))=h(x)=0$, hence we can write $g(w)=i'(v')$, and $\delta(x):=[v']$.
Hence, $\delta$ is trivial iff for every $w \in W$ we have: If $p'(g(w))=0$, then $g(w)=i'(f(v))$ for some $v$. In other words, $im(g) \cap im(i')$ equals $im(i'f)$, which means exactness as explained above.
