Induced Exact Sequence of Dual Spaces 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.
 A: The exactness of the first sequence means that $S$ is injective, $T$ surjective, and the range of $S$ meets the kernel of $T$ just the right way in $V$.
Okay, so to show that the second sequence is exact, we'll start by showing $\circ T$ is injective. Let $g,g'$ be elements of $W^{*}$. Suppose that $g(T) = g'(T)$. Since $T$ is surjective, for any $w \in W$ there is some $v \in V$ so that $T(v) = w$. Then $g(T(v)) = g'(T(v))$ so that $g(w) = g'(w)$, so that $g$ and $g'$ are the same on all elements of $W$, and hence are the same element of $W^*$.
Next, we'll show that $\circ S$ is surjective. Let $h$ be an arbitrary element of $U^*$. We want to produce an element $f \in V^*$ such that $f(S) = h$. We can define $f$ on the range of $S$, knowing that it can be extended to a linear functional on all of $V$. On the range of $S$, define $f$ to be $h(S^{-1})$. This makes sense, since $S$ is injective. Then $f(S) = h(S^{-1}(S)) = h$, proving surjectivity of $\circ S$.
I'll leave it to you to verify that $V^*$ splits as the range of $\circ T$ and the kernel of $\circ S$, using the techniques outlined in the prior steps.
A: Suppose your vector spaces are all over some field $\Bbb{F}$. Now the functor $\textrm{Hom}(-,\Bbb{F})$ in general is only right exact. However because $\Bbb{F}$ as a module over itself is injective, the functor $\textrm{Hom}(-,\Bbb{F})$ is exact and so we get the exact sequence 
$$0 \longrightarrow \textrm{Hom}(W,\Bbb{F}) \longrightarrow \textrm{Hom}(V,\Bbb{F}) \longrightarrow \textrm{Hom}(U,\Bbb{F}) \longrightarrow 0.$$
Or you can notice that in the long exact sequence of Ext groups, the boundary map
$$\partial : \textrm{Hom}(U,\Bbb{F}) \rightarrow \textrm{Ext}^1_{\Bbb{F}}(W,\Bbb{F})$$
is actually the zero map because $\textrm{Ext}^1_{\Bbb{F}}(W,\Bbb{F}) = 0$. Hence $\textrm{Hom}(-,\Bbb{F})$ is an exact functor which completes the problem.
A: 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 $\mathrm{Ker}(S^*) \subseteq \mathrm{Im}(T^*)$. 
A: It follows from the fact, that all $Ext^{>0}$ groups are zero over a field.
Then write the long exact sequence.
