Precisely
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Associativity of the tensor product

All vector spaces are by default over \K = \R or \C.

We consider three vector spaces U, V and W. Let (P, \alpha) be a tensor product of U and V, and (Q, \beta) a tensor product of P and W.

The aim is first to show that Q, together with the mapping:

    \[\gamma: U \times V \times W \to Q, (\vec u, \vec v, \vec w) \mapsto \beta(\alpha(\vec u, \vec v), \vec w)\]

is a tensor product of U, V and W.

Proof:

We note that \gamma as defined above is trilinear.

Let Q' be a vector space and \gamma' a trilinear mapping U \times V \times W \to Q'. We must show that there exists a unique linear mapping \lambda: Q \to Q' such that \lambda \circ \gamma = \gamma'.

For any \vec w \in W, the mapping U \times V \to Q', (\vec u, \vec v) \mapsto \gamma'(\vec u, \vec v, \vec w) is bilinear; hence there is a unique linear mapping \beta'_{\vec w}: P \to Q' such that for all (\vec u, \vec v) \in U \times V, we have \gamma'(\vec u, \vec v, \vec w) = (\beta'_{\vec w} \circ \alpha)(\vec u, \vec v).

It is easy to check that the mapping P \times W \to Q', (\vec p, \vec w) \mapsto \beta'_{\vec w}(\vec p) is bilinear; hence there is a unique linear mapping \lambda_0: Q \to Q' such that for all (\vec p, \vec w) \in P \times W, we have (\lambda_0 \circ \beta)(\vec p, \vec w) = \beta'_{\vec w}(\vec p).

Then for any (\vec u, \vec v, \vec w) \in U \times V \times W, with \vec p = \alpha(\vec u, \vec v) \in P:

(\lambda_0 \circ \gamma)(\vec u, \vec v, \vec w) = \lambda_0(\gamma(\vec u, \vec v, \vec w)) = \lambda_0(\beta(\alpha(\vec u, \vec v), \vec w))= (\lambda_0 \circ \beta)(\vec p, \vec w) = \beta'_{\vec w}(\vec p) = \gamma'(\vec u, \vec v, \vec w)

Hence \lambda = \lambda_0 is such that \lambda \circ \gamma = \gamma'.

We must now show that if \lambda: Q \to Q' is such that \lambda \circ \gamma = \gamma', then \lambda = \lambda_0.