Tensor Product of Vector Space and Field Please let me know if the solution to the following question seems correct.
Let $V\in  \operatorname{Vect}(k)$.  Show $k\otimes V\cong V$
Solution----------------
Let $c\in k$, $v\in V$ and $f:k\times V\to V$ be defined as $f(c,v)=cv$.  Note, $f$ is bilinear.  Let $W\in \operatorname{Vect}(k)$, $w\in W$ and suppose $g:k\times V\to W$ is also bilinear.  Since $g$ is bilinear we have $g(c,v)=g(1,cv)\in V$ hence $V\cong W$ and the diagram  
$$
\require{AMScd} 
\begin{CD}
        k\times V @>f>> V \\
         @VgVV @VVIV \\
        W @= W
    \end{CD}
$$
commutes.  Since the identity is unique we have that $V$ is the tensor product of $k\otimes V$ giving us the result.
 A: You are on the right track with $f$ defined as you did. It seems to me that you want to show that $V$ satisfies the universal property of tensor product. This works, but you didn't really write down all the necessary details.
First of all, tensor product of $A$ and $B$ is not just some vector space $C$. It comes with bilinear map $\otimes\colon A\times B\to C$ as well. In your case, $f$ plays the role of $\otimes$.
Now, what you want to show is that for each bilinear $g\colon k\times V\to W$, there exists unique $\bar g\colon V\to W$ such that $\bar g\circ f = g$. You didn't properly define $\bar g$ nor show its uniqueness. It looks like $I$ in your diagram should play the role of $\bar g$, but you seem to think that this is identity, if I understood correctly. It is not identity and $V$ is not isomorphic to $W$ in general. $W$ should be arbitrary vector space.
So, how would we define $\bar g$? You wrote yourself that $g(c,v) = g(1,cv)$, but we should go one step further, i.e. $g(c,v) = g(1,f(c,v))$ and this tells you what your $\bar g$ should be: $\bar g (v)=g(1,v)$. Obviously it is linear and $g(c,v) = g(1,f(c,v)) = \bar g(f(c,v))$.
What now remains is uniqueness. So, let $h\colon V\to W$ be a linear map such that $h\circ f = g$. Then, $h(v) = h(f(1,v)) = g(1,v) = \bar g(v)$ for all $v\in V$, and we are done.
Since $(V,f)$ satisfies the universal property of tensor product, by uniqueness of tensor product up to unique isomorphism there is unique isomorphism $\varphi\colon k\otimes V\cong V$ such that $\varphi\circ\otimes = f$. It is easy to see that $\varphi = \bar f$, the map that $f$ induces by the universal property of $k\otimes V$.

There is much shorter way to go about this. What you want to exhibit is map $g\colon V\to k\otimes V$ that is inverse of $\bar f\colon k\otimes V\to V$ induced by the universal property of tensor product.
Hint: What is the simplest non-zero map $g\colon V\to k\otimes V$ you can think of?
