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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.

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    You can do this very explicitly by choosing appropriate bases of $U, V, W$ and looking at the corresponding dual bases.2012-10-29
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    @QiaochuYuan: Is there then a problem in the case that these spaces aren't finite dimensional? Isn't the dual set of a basis $\mathcal{B} \subseteq V$ a basis for $V^*$ iff $\text{dim}(V)\lt \infty$?2012-10-29
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    I missed that you didn't want to restrict $U, V, W$ to be finite-dimensional.2012-10-29
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    If I remember correctly, Greub proves this on his _Linear Algebra_ book.2018-05-30

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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.

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    Interesting approach. And forgive my ignorance here. But how do these show exactness at each stage? Aren't the things you're proving necessary consequences if we presuppose exactness? Or do they also imply exactness?2012-10-29
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    Isn't this the definition of exactness?2012-10-29
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    If it is, I can't see its equivalence to the one I know! Which may be my fault. The one I'm familiar with is $\text{im}(S)=\text{ker}(T)$.2012-10-29
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    Yes, these definitions are equivalent. To see this, observe that the image of $0 \to W^*$ is zero, so that must be the kernel of $W^* \to V^*$. But saying that the kernel is zero is the same thing as saying that a map is injective. Similarly, I've shown that the map $V^* \to U^*$ is surjective, which means that its image is all of $U^*$. Since the kernel of $U^* \to 0$ is also all of $U^*$, we see that this is same as regular exactness.2012-10-29
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    Awesome. I get it! Now how does proving that $V^*$ splits fit in? Or is that something extra?2012-10-29
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    It's something extra I've left to you. Try working on it, and if you don't have any luck I'll include it above.2012-10-29
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    @IsaacSolomon Sorry for bringing this back again, but how does this prove that the kernel of $S^*$ equals the image of $T^*$? you have regular exactness on the extremes, but not in the middle.2015-05-25
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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.

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

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    Since I've verifiable noob status, I have no idea what this means!2012-10-29
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    You can read it in Wikipedia: http://en.wikipedia.org/wiki/Ext_functor2012-10-29
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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 $\mathrm{Ker}(S^*) \subseteq \mathrm{Im}(T^*)$.