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?
$\bf{\text{Context:}}$
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 $\otimes_{i\in I}X_{i}$.
Best Answer
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.