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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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