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In my homework, it asks me to show that the operation $-\otimes_{k} V$ sends exact sequence to exact sequence. What does the operation mean in terms of the map? For example if the map from $C_{n}$ to $C_{n+1}$ was $c \mapsto c'$, does the new map looks like $c \otimes v \mapsto c' \otimes v$ (i.e. the $v$ stays the same)?

BetaY
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  • Your description of the action on $c\otimes v$ is correct. (For a general element of $C_n\otimes V$, which will be a sum of terms like $c\otimes v$, you'll want to also use the fact that the map you want is linear.) – Andreas Blass Oct 19 '15 at 09:35

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Yes, that's exactly it. More formally: the functor $- \otimes_k V$ sends a $k$-vector space $A$ to $A \otimes_k V$. Given a linear map $f : A \to B$, the map $A \times V \to B \otimes_k V$ given by $(a,v) \mapsto f(a) \otimes v$ is $k$-bilinear, and hence induces a linear map $A \otimes_k V \to B \otimes_k V$. This is the map you want.

(Just a remark, but as usual you have to be careful when you define maps like that: sometimes it's not bilinear, and the map doesn't exist. For example it seems reasonable at a glance to write something like "$V \otimes_k V \to V$, $v \otimes v' \mapsto v + v'$"... But the map I've written is not bilinear, so this doesn't exist!)

Najib Idrissi
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  • To prove this operation sends exact sequence to exact sequence, is it true that ker$(d_n ')=$ker$(d_n) \otimes V$, where $d_n'$ is the new map? – BetaY Oct 19 '15 at 13:33
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    This is basically what you need to prove, @BayLee. It's not so easy (hint: use the fact that you're working with vector spaces). Either search for "tensor product exactness", "vector space flatness" or something similar; or ask a new question if you can't find what you need, though I'm pretty sure this has been asked before. – Najib Idrissi Oct 19 '15 at 13:39