diagram of short exact sequence I have this commutative diagram of vector complex spaces where all the sequences that appear are short and exact.

is there a way to say that $H$ is the intersection between $W1$ and $W2$?
 A: A much simpler diagram suffices (or the symmetrical subdiagram with $Z$ instead of $T$):
$$\begin{matrix}
&&&0\\
&&&\downarrow\\
0\to&H&\to &W_1&\to&T\\
&\downarrow&\searrow&\downarrow\rlap{\scriptstyle\subseteq}&&\|\\
0\to &W_2&\stackrel\subseteq\to &V&\to &T
\end{matrix} $$
where


*

*all meshes commute,

*the middle column is exact at $W_1$ (which just rephrases that we view $W_1$ as subspace of $V$)

*the top row is exact 

*the lower row is a complex and is exact at $W_2$ (again this just rephrases that we view $W_2$ as subspace of $V$)


Assume $w\in W_1\cap W_2$. Then $w\mapsto 0$ under $W_2\to V\to T$, hence also under $W_1\to V\to T$, hence by exactness of the top row $w$ "comes" from $H$. On the other hand, the diagonal arrow maps any element of $H$  to an element of both $W_1$ and $W_2$ (namely along $H\to W_1\hookrightarrow V$ and $H\to W_2\hookrightarrow V$). The found maps $H\leftrightarrow W_1\cap W_2$ are inverse.
A: If you think $H,W_1,W_2$ as subspaces of $V$, then you have $H = W_1 \cap W_2$.
Clearly $H \subseteq  W_1 \cap W_2$ (simply look at the diagram).
Viceversa, if $w \in W_1 \cap W_2$, then you have
$$w \in \ker (V \longrightarrow T) \cap \ker (V \longrightarrow Z)$$
so
$$w \in \ker (W_1 \longrightarrow T) \cap \ker (W_2 \longrightarrow Z) = \operatorname{Im} (H \longrightarrow W_1) \cap \operatorname{Im} (H \longrightarrow W_2) = H$$
