I want to show that $X\otimes(Y\otimes Z)$ is isomorphic to $(X\otimes Y)\otimes Z$.

Intuitively I think I should just choose bases $\{e_{i}\}_{i\in I}, \{f_{j}\}_{j\in J}$, and $\{g_{k}\}_{k\in K}$ for $X,Y,Z$ and map

$$e_{i}\otimes(f_{j}\otimes g_{k})\mapsto (e_{i}\otimes f_{j})\otimes g_{k}$$

This defines a bijective correspondence between basis elements and so should induce a vector space isomorphism.

Is there a way to do this using the universal property of the tensor product?


I am working through Introduction to Tensor Products of Banach Spaces by Raymond Ryan, and am working on exercises at the end of the first chapter. I'll try to summarize what I have available.

We defined the space $X\otimes Y$ to be linear functionals on the space $B(X\times Y)$ of bilinear maps on $X\times Y$.

That is, $x\otimes y:A\mapsto A(x,y)$ for $A\in B(X\times Y)$.

We have stated and proved

$\bf{\text{Universal Property of Tensor Products}}$: Let $X,Y,Z$ be vector spaces. For every bilinear $A:X\times Y\to Z$ there is a unique linear map $\hat{A}:X\otimes Y\to Z$ such that $\hat{A}(x\otimes y) = A(x,y)$.

Next we proved that the Tensor product is unique up to isomorphism (in the sense of having this property).

We did not define any higher tensor product structure such as $\otimes_{i\in I}X_{i}$.

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    $\begingroup$ Both are isomorphic to $X\otimes Y\otimes Z$, which represents multilinear maps out of $X\times Y\times Z$. $\endgroup$ – wildildildlife Jul 6 '13 at 20:41
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    $\begingroup$ Don't I need to first show associativity before I can even make sense of the expression $X\otimes Y\otimes Z$? (More details on the context added to question momentarily.) $\endgroup$ – roo Jul 6 '13 at 20:49
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    $\begingroup$ First Step: Show that, for every $x$, the mapping $(y,z)\mapsto(x\otimes y)\otimes z$ is bilinear. $\endgroup$ – Stefan Jul 6 '13 at 20:57
  • $\begingroup$ Got it. I linearize that map into a map $A_{x}:Y\otimes Z\to (X\otimes Y)\otimes Z$. Then I linearize the map $(x,y\otimes z)\mapsto A_{x}(y\otimes z)$ to get the isomorphism. Thanks Stefan! $\endgroup$ – roo Jul 6 '13 at 21:17
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    $\begingroup$ That's right! Don't forget, however, that not every element of $Y\otimes Z$ is of the form $y\otimes z$. $\endgroup$ – Stefan Jul 6 '13 at 21:48

Here's a sketch of the argument based on the hint from Stefan Walter.

Start by fixing $x\in X$, and define the bilinear map $Y\times Z\to (X\otimes Y)\otimes Z$ by $$(y,z)\mapsto (x\otimes y)\otimes z$$

By the universal property stated in my question, this induces a linear map $$A_{x}:Y\otimes Z\to (X\otimes Y)\otimes Z\text{ such that }A_{x}(y\otimes z) = (x\otimes y)\otimes z\text{ for every }y\in Y,z\in Z$$

Next define the bilinear map $X\times (Y\otimes Z)\to (X\otimes Y)\otimes Z$ by $$(x,\sum_{i=1}^{n}y_{i}\otimes z_{i})\mapsto A_{x}(\sum_{i=1}^{n}y_{i}\otimes z_{i})$$

Passing this map through the universal property yields the isomorphism.


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