My question:
Take a manifold $Z$, and let $S\subset Z$ be a submanifold of $Z$ (with the subspace topology).
Let $V$, $V'$, and $V''$ be vector bundles on $Z$, with projections $\pi:V \to Z$ etc.
Write $V|_S$ for the restriction of the bundle $V$ to $\pi^{-1}(S)$, with the restricted projection $\pi_S: V|_S \to S$. That is, $V|_S$ is the pullback bundle $i^* V$, where $i: S \to Z$ is the inclusion. Then $V|_S$ is a vector bundle over $S$. Similarly for $V'|_S$ and $V''|_S$.
Suppose now that we have a vector bundle map $u:V \to V'$. That is, suppose $u: V \to V'$ is a continuous map and that the following diagram commutes, $$ \newcommand{\ra}[1]{\,\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\,\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} V & \ra{~~~u~~~} & V' \\ \da{\pi} & & \da{\pi'} \\ Z & \ras{~~~\text{Id}~~~} & Z \\ \end{array} $$
I believe this induces a vector bundle map $\tilde{u}:V|_S \to V'|_S$, simply by restricting all four maps. (More generally, I imagine it is the case that a vector bundle map $u: E \to E'$, where $E$ and $E'$ are vector bundles over $X$ and $Y$, induces a vector bundle map $\tilde{u} : f^*E \to f^*E'$, for any continuous map $f: X \to Y$ between the base spaces. But I have not seen a proof of this.)
Now suppose we have a short exact sequence $$0 \to V \xrightarrow{\,f\,} V' \xrightarrow{\,g\,} V'' \to 0$$ of vector bundles on $Z$. My question is whether this induces a short exact sequence for the vector bundles on $S\subset Z$, $$0 \to V|_S \xrightarrow{\,\tilde{f}\,} V'|_S \xrightarrow{\,\tilde{g}\,} V''|_S \to 0 \, .$$
My reasoning:
I believe that the answer to this question is affirmative, and my reasoning is as follows. Please let me know if anything here is incorrect.
Since $\tilde{f}$ is the restriction of an injective map $f$, it too is injective.
As $g$ is surjective, and since points in $\pi''^{-1}(S)$ can only be mapped to from $\pi'^{-1}(S)$ (because of the commutative diagram condition, which enforces that a point in a fibre over a point $x\in Z$ in the base is mapped to a point in a fibre over the same point), $\tilde{g}:\pi'^{-1}(S)\to\pi''^{-1}(S)$ is also surjective.
To show the sequence is exact, all that remains is to show $\text{Im}(\tilde{f})=\text{Ker}(\tilde{g})$. For this we can note that $$\text{Im}(\tilde{f})=\text{Im}(f)\,\cap\,\pi'^{-1}(S)\, ,$$ $$\text{Ker}(\tilde{g})=\text{Ker}(g)\,\cap\,\pi'^{-1}(S)\, .$$ (The second is I think obvious. The first is perhaps more difficult to see, but again seems to follow from the fact that a fibre over $x\in Z$ is mapped to another fibre over the same point.) Hence since $\text{Im}(f)=\text{Ker}(g)$, we have that $\text{Im}(\tilde{f})=\text{Ker}(\tilde{g})$.
So it seems that we have induced a short exact sequence on the subspace $S$.
My specific interest:
In particular, I'm interested in showing that a short exact sequence $0 \to V \to V' \to V'' \to 0$ on an elliptic fibration $Z$ induces a short exact sequence $0 \to V|_{E_b} \to V'|_{E_b} \to V''|_{E_b} \to 0$ on each elliptic curve $E_b$ ($b\in Z$).
More specifically, I would like to show that if we begin with a non-trivial extension on $Z$ then we get non-trivial extensions on each elliptic curve $E_b$.