Tensor product of function space and vector space In one book I found that they treat vector valued functions as tensor product of some function space and vector space. So for example
$$
C(\mathbb{R},V) \simeq C(\mathbb{R})\otimes V \\
L^p(\mathbb{R},V) \simeq L^p(\mathbb{R})\otimes V
$$
Suppose I define tensor product through universal property in some category.
The question is, which category should I choose? And how do I prove that for example $C(\mathbb{R},V) \simeq C(\mathbb{R})\otimes V $ is true ?
My guess is that I should choose category of Banach spaces, but I'm not sure how should I go about proving the rest.

Connection to Bochner integral.
It seams to me that in this way I can define Bochner integral very easily. If I take map($B$ is some Banach space)
$$
F: L^1(\mathbb{R})\times B \rightarrow B\\
F(f,v) = v\int_\mathbb{R} f
$$ 
Than $F$ uniquely extends to $L^1(\mathbb{R})\otimes B \simeq L^1(\mathbb{R},B)$. It gives the right answer to the integral of simple function, so I guess it should be the Bochner integral.
Is there a problem in defining the integral this way?
 A: The tensor product of vector spaces is defined via its usual universal property. We then get, for every finite-dimensional real vector space $V$ and every topological space $X$, a canonical linear map $C(X,\mathbb{R}) \otimes V \to C(X,V)$ mapping $f \otimes v$ to $(x \mapsto f(x) v)$. In order to make sense of the right hand side, we have to recall that $V$ carries a unique topology which makes $V$ a topological vector space. It is easy to see that the linear map is actually an isomorphism. Since both sides are additive in $V$, it suffices to prove this for $V=\mathbb{R}$, where it is trivial.
When $V$ is an arbitrary topological vector space, then we can still define the linear map but it won't be an isomorphism anymore. It has a good chance to be injective, but not chance to be surjective. Instead, one has to complete the tensor product (various completions are available). If $V$ is a Banach space and $X$ is a compact Hausdorff space, then I think that $C(X,\mathbb{R}) \widehat{\otimes} V \cong C(X,V)$ holds. Perhaps someone else can add a reference?
$L^1(X,V)$ makes sense if $X$ is a measurable space and $V$ is a Banach space. Again there is a canonical linear map $L^1(X,\mathbb{R})  \otimes V \to L^1(X,V)$ which should (?) extend to an isomorphism $L^1(X,\mathbb{R}) \widehat{\otimes} V \cong L^1(X,V)$.
And yes, this way you can reduce the Bochner integral to the $1$-dimensional integral, but I think the construction of the Bochner integral is as easy as the $1$-dimensional case, in fact the only difference is that you replace the absolute value of $\mathbb{R}$ by the norm on any Banach space $V$, which doesn't cause any difficulties.
