Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You can do this very explicitly by choosing appropriate bases of $U, V, W$ and looking at the corresponding dual bases. – Qiaochu Yuan Oct 29 '12 at 6:23
@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 $\text{dim}(V)\lt \infty$? – AsinglePANCAKE Oct 29 '12 at 14:14
I missed that you didn't want to restrict $U, V, W$ to be finite-dimensional. – Qiaochu Yuan Oct 29 '12 at 17:28

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.

share|cite|improve this answer
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? – AsinglePANCAKE Oct 29 '12 at 5:46
Isn't this the definition of exactness? – Isaac Solomon Oct 29 '12 at 13:39
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 $\text{im}(S)=\text{ker}(T)$. – AsinglePANCAKE Oct 29 '12 at 14:19
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. – Isaac Solomon Oct 29 '12 at 17:08
Awesome. I get it! Now how does proving that $V^*$ splits fit in? Or is that something extra? – AsinglePANCAKE Oct 29 '12 at 17:12

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.

share|cite|improve this answer

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

share|cite|improve this answer
Since I've verifiable noob status, I have no idea what this means! – AsinglePANCAKE Oct 29 '12 at 6:09
You can read it in Wikipedia: – user46336 Oct 29 '12 at 6:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.