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I have seen two different definitions of the tensor product and I am asking about their relation.

  1. Definition (for modules in most abstract algebra books). Given two vector spaces $U$ and $V$ over $F$, the tensor product $T = U \otimes V$ is a vector space $T$ over $F$ together with a bilinear map $\pi : U \times V \to T$ such that every bilinear map on $U$ and $V$ factors through $T$ (by a unique linear map). This could easily be extended to the tensor product of an arbitrary but finite number of vector spaces.

  2. Definition (from Spivak, Calculus on Mainfolds and Munkres, Analysis on Mainfolds). Given a vector space $V$ over $F$, the $k$-fold tensor is the set of all multilinear maps from $V^k$ to $F$ (hence the tensor product of two spaces is the set of bilinear maps).

    What is the relation of Definition 1 and Definition 2?

The first definition gives an isomorphism $\operatorname{Bilin}(U, V, F) \cong L(U \otimes V, F)$ as the correspondence giving the unique linear map is linear. In general this gives an isomorphism between the $k$-fold multilinear maps and the linear maps from the $k$-fold tensor product to $F$ (note that we have not used the property to its full extend, as it also holds for bilinear maps not having their image in $F$, but lets restrict to that case).

But still this does not identifies the tensor product itself with the multilinear maps, but just those maps with the dual of the tensor product (which might in general not equal the tensor product itself)?

When we start from Definition 2, I am not quite sure how to show the universal property from Definition one. So lets consider the $2$-fold tensor product of $V$ over $F$, then we have to show that $\operatorname{Bilin}(U,V, F)$ fulfills this (universal) property. The only possible way that comes to my mind is to define $$ \pi(u,v) = u^{\ast} \cdot v^{\ast} $$ where $u^{\ast}$ and $v^{\ast}$ denote the correspoding dual elements of $V^{\ast}$ (after choosing a basis in $V$ and thereby having a dual basis in $V^{\ast}$ according to the usual construction, see here). And if $\varphi : U \times V \to F$ is another bilinear map, we have to show that $$ \varphi(u,v) = h(u^*\cdot v^*) $$ for a unique linear map, if we define $h$ that way (using that $u \mapsto u^{\ast}$ in injective) we could show that it is linear (using $\{ e_i^{\ast}\}$ and some computations). But in general $\{ u^{\ast} \mid u \in V \}\ne V^{\ast}$, so it is not clear to me how to extend $h$ to all of $\mbox{Bilin}(U,V,F)$, hence this only works in the case of reflexivity (which include the finite-dimensional vector spaces).

So what I have done just works for reflexive vector spaces, but am I on the right track? Maybe there are simpler arguments, and Munkres does not restrict his definition to finite-dimensional vector spaces only, so does it makes sense to define the tensor product like in definition two, but then how does it relate to definition one?

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Definition 2 is actually the dual of the tensor product of $n$ copies of $V$. One can show that if $V$ is finite dimensional, then one has a canonical isomorphism $$(V\otimes\cdots\otimes V)^*\cong V^*\otimes\cdots\otimes V^*.$$ If moreover we fix a basis of $V$ (or more generally an inner product), then we have a canonical isomorphism $V\cong V^*$, so that the two definitions coincide. However, the "correct" definition is the first one.

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