Tensor product of an exact sequence of vector spaces by a vector space Can you give a simple proof that tensoring on the right by a vector space, preserves exactness of given exact vector spaces? (with the obvious identity map for the tensor multiplied on the right) I have seen proofs for general cases but for vector spaces, is there an easy proof? 
 A: Let $V$ be a vector space over a field $F$ with basis $v_i : i \in I$.  If $W$ is another vector space with basis $w_j : j \in J$, then $V \otimes W$ is a vector space with basis $v_i \otimes w_j, (i,j) \in I \times J$.  It follows immediately from here that if $v_1, ... , v_n$ are linearly independent in $V$, and $a_1, ... , a_n$ are nonzero elements of $W$, then $v_1 \otimes a_1, ... , v_n \otimes a_n$ is a linearly independent set in $V \otimes W$.
If $f: W_1 \rightarrow W_2$ is a linear transformation, then $$1_V \otimes f: V \otimes W_1 \rightarrow V \otimes W_2$$ is the linear transformation given on generators by $v \otimes w_1 \mapsto v \otimes f(w_1)$.
Now let $$W' \xrightarrow{f} W \xrightarrow{g} W''$$ be an exact sequence of vector spaces.  We need to show that 
$$V \otimes W' \xrightarrow{1 \otimes f} V \otimes W \xrightarrow{1 \otimes g} W''$$ is exact.  It is immediate that $[1 \otimes g] \circ [1 \otimes f] = 0$.  Now we just need to show that if $x \in V \otimes W$, and $[1 \otimes g](x) = 0$, then $x$ lies in the image of $ 1 \otimes f$.  Write $x$ as 
$$x = \sum\limits_{i=1}^n v_i \otimes y_i$$ where $v_1, ... , v_n$ are linearly independent in $V$, and $y_1, ... , y_n$ are elements of $W$.  Then
$$ 0 = [1 \otimes g](x) = \sum\limits_{i=1}^n v_i \otimes g(y_i)$$ 
By the remark in the first paragraph I wrote, applied to $W''$ instead of $W$, the only way the above sum can be zero is if all the coefficients $g(y_i)$ are zero.  By the exactness of our original sequence, this implies that $y_i = f(z_i)$ for some $z_i \in W'$.  Thus 
$$x = \sum\limits_{i=1}^n v_i \otimes y_i = \sum\limits_{i=1}^n v_i \otimes f(z_i) = [1 \otimes f](\sum\limits_{i=1}^n v_i \otimes z_i)$$
