Induced Exact Sequence of Dual Spaces

Question

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.

Answer 1

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.

Answer 2

It follows from the fact, that all $Ext^{>0}$ groups are zero over a field. Then write the long exact sequence.

Answer 3

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.