TITLE: Induced Exact Sequence of Dual Spaces
QUESTION [8 upvotes]: So given a short exact sequence of vector spaces $$0\longrightarrow U\longrightarrow V \longrightarrow W\longrightarrow 0$$  With linear transformations $S$ and $T$ from left to right in the non-trivial places.
I want to show that the corresponding sequence of duals is also exact, namely that $$0\longleftarrow U^*\longleftarrow V^* \longleftarrow W^*\longleftarrow 0$$
with functions $\circ S$ and $\circ T$ again from left to right in the non-trivial spots.  So I'm a bit lost here.  Namely, I'm not chasing with particular effectiveness.  Certainly this "circle" notation is pretty suggestive, and I suspect that this is a generalization of the ordinary transpose, but I'm not entirely sure there either.
Any hints and tips are much appreciated.

REPLY [5 votes]: You can actually prove a more general result: that if you have a sequence $U \xrightarrow{S} V \xrightarrow{T} W$ with the property that $\mathrm{Ker}(T) = \mathrm{Im}(S)$, i.e. the sequence is exact at $V$, then the dual sequence $W^* \xrightarrow{T^*} V^* \xrightarrow{S^*} U^*$ is also exact at $V^*$; that is, $\mathrm{Ker}(S^*) = \mathrm{Im}(T^*)$. 
Indeed, if $f \in W^*$ and $u \in U$, then $$(S^* \circ T^*)(f)(u) = T^*(f)(S(u)) = f((T \circ S)(u)) = 0$$ because $T \circ S = 0$ by exactness of $V$. So $\mathrm{Im}(T^*) \subseteq \mathrm{Ker}(S^*)$. 
On the other hand, suppose $f \in \mathrm{Ker}(S^*)$. This means for every $u \in U$, $S^*f(u) = f(S(u)) = 0$. We want a $g \in W^*$ so that $T^* g = f$. So define $g \in \mathrm{Im}(T)^*$ by $g(T(v)) = f(v)$. This is well-defined on $\mathrm{Im}(T) \subseteq W$, because if $T(v) = T(v')$ for $v, v' \in V$, then $T(v-v') = 0$, so $v-v' \in \mathrm{Ker}(T) = \mathrm{Im}(S)$, so there is a $u \in U$ so that $S(u) = v - v'$; but since $f \in \mathrm{Ker}(S^*)$, this means $g(T(v-v')) = f(v-v') = f(S(u)) = 0$, so $g(T(v)) = g(T(v'))$. Let $\tilde W$ be any subspace of $W$ so that $W = \mathrm{Im}(T) \oplus \tilde W$, and declare $g|_{\tilde W} = 0$. Then $g \in W^*$ satisfies $T^*g(v) = g(T(v)) = f(v)$ for all $v \in V$, so $T^* g = f$, so $\mathrm{Ker}(S^*) \subseteq \mathrm{Im}(T^*)$.