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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.

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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?

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    Great answer. I'll come back and work on your simpler version later. – 2017-01-14
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    This is a minor issue but where you write $\bar g (v)=g(1,v)$ shouldn't it actually be $\bar g (cv)=g(1,cv)$. Or maybe $\bar g (v^{\prime})=g(1,v^{\prime})$ for any $v^{\prime}\in V$ – 2017-01-14
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    In regard to your hint is $g$ the map taking $v\in V\mapsto 1_k\otimes v$? – 2017-01-14
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    Well, the correct way would be $(\forall v\in V)\ \bar g(v) = g(1,v)$, which is exactly what you write with $v'$. But there really is no reason to change $v$ to $v'$, all the stuff where quantification is not used (and should be) is to be interpreted as universal quantification. On the other hand, $\bar g(cv) = g(1,cv)$ is exactly the same as the other stuff, since $v\mapsto cv\colon V\to V$ is isomorphism for all $c\neq 0$. – 2017-01-14
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    And, yes, $g(v) = 1_k\otimes v$ is the required map. You just have to show that $g^{-1} = \bar f$ and you have your isomorphism, @Bob. – 2017-01-14
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    One last question, I just want to confirm that $\bar f(cv)=cv$ – 2017-01-16
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    It is not, that would be identity. $cv\not\in k\otimes V$, instead $\bar f(c\otimes v) = f(c,v) = cv$, @Bob. – 2017-01-16