I found the following question in an online book of Linear Algebra exercises (page 69)
All of the mathematical objects are vector spaces, and thus the maps are linear. It does not specify if the vector spaces are finite dimensional, which I don't think makes a difference here (see my work below). The book also states the unlabeled arrows are the "obvious" linear maps. The left most would be the map from the vector $0$ to $0$, and the right most would be the zero map on its domain.
Point (h), the problem to solve, seems a little ambiguous, and it is not phrased as a question. However, I am assuming the statement is true and it is asking me to prove it.
The issue is, I don't see why $h$ must be surjective if $g$ is surjective, for the following reason:
$\quad$ Because the left square commutes, $j' \circ f = g \circ j$, since
$g$ is surjective, so must be $j'$. In fact $j'$ is invertible
since it is also injective, since the bottom row is exact.
$\quad$ Since $j'$ is surjective, and the bottom row is exact: $ V' =
range(j') = ker(k') \implies k' = 0_{V'}$. Thus by the right square commuting, $range(h' \circ
k) = {\bf{0}}$. But since the top sequence is exact, $j$ is
injective $\implies$ $k$ is surjective. Since $k$ is surjective, and $range(h' \circ
k) = {\bf{0}}$ , means $h = 0_W$. So
$h$ is the zero map and can't be surjective. Thus it looks like the statement isn't true.
Also it seems no Linear Algebra is actually needed.
I'm not sure if there is something I am missing. I would appreciate any hints.
Best Answer
Your reasoning for $j'$ being surjective is simply incorrect.
For any composition of maps $hk$ we have that the image of $hk$ is contained in the image of $h$.
If we assume $g$ is surjective, then since $k'$ is surjective $k'g = hk$ is surjective. This means the image of $hk$ is all of $W'$ and so $h$ must be surjective as well.