1

Given a long exact sequence of vector spaces: $$...\longrightarrow V_1 \overset{f}{\longrightarrow}V_2\overset{g}{\longrightarrow}V_3\longrightarrow...$$ Given another vector space $W$, is the following sequence still exact?How to write down the map between them explicitly?

$$...\longrightarrow V_1\otimes W \overset{?}{\longrightarrow}V_2\otimes W\overset{?}{\longrightarrow}V_3\otimes W\longrightarrow...$$

There are few books on homological algebra at my hand but I haven't found any detailed proof for this.

Any hints or book recommendation will be appreciated!

C Weid
  • 660
  • Not necessarily. The tensor product is right exact but not left exact, and a long sequence can equivalently be written as a bunch of short exact sequences. All of this can be found in Weibel's book, for example. – Najib Idrissi Aug 17 '14 at 12:17
  • The question is about vector spaces over a field. Everything is flat in this world. – apurv Aug 17 '14 at 13:16

2 Answers2

1

The statement is true for vector spaces but not for arbitrary modules.

Here are some steps, you can try filling in the gaps:

Show that it is enough to prove that if $0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0 $ is exact then $0 \rightarrow V_1 \otimes W \rightarrow V_2\otimes W \rightarrow V_3\otimes W \rightarrow 0 $ is exact.

Next show that for this it is enough to prove that $ (V_1 \oplus V_2) \otimes W \cong (V_1 \otimes W) \oplus (V_2 \otimes W) $ canonically.

Finally use bases and construct an explicit isomorphism.

apurv
  • 954
0

I'm not an expert on homological algebra, so maybe there is a more elegant way of doing this but one possibility should be:

First decompose you sequence into short exact sequences (this is possible e.g. because if you have a map of vector spaces, you can decompose your source space as direct sum of kernel and image)

Show that the tensor product is always right exact (i.e. it preserves surjective maps and also exactness in the middle) this is a general fact about modules and it's not necessary to use anything specific on vector spaces, but maybe there is a faster way if you do.

Show that also injectivity is preserved (use for example the fact that you can write down a basis of the tensor product space as products of basis elements of the two factor spaces)

How does the map look like: It is the same map on the left factor of every elementary tensor and the identity on the right.

It is a bit sketchy, so if you get stuck anywhere feel free to ask for more :)

jorst
  • 1,555