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This question already has an answer here:

Prove $(U \otimes V) \otimes W \cong U \otimes (V \otimes W)$.

$U, V, W$ are finite dimensional vector spaces over a field. How can I prove that using the universal property of the tensor product?

Is there an isomorphism $\phi: (U \otimes V) \otimes W \to U \otimes (V \otimes W)$ that satisfies $\phi((u \otimes v) \otimes w) = u \otimes (v \otimes w)$?

Thank you in advance.

Edit: Is this a valid proof?

Let $(e_1, ..., e_n)$ be a basis of $U$, $(f_1, ..., f_m)$ be a basis of $V$, $(g_1, ..., g_r)$ be a basis of $W$. Then $(((e_i \otimes f_j), w_k))_{1 \leq i \leq n, 1 \leq j \leq m, 1\leq k \leq r}$ is a basis of $(U \otimes V) \times W$.

Define $\phi: (U \otimes V) \times W \to U \otimes (V \otimes W)$ by $\phi((e_i \otimes f_j), w_k) := e_i \otimes (f_j \otimes g_k)$. $\phi$ is bilinear.

Claim: $(U \otimes (V \otimes W), \phi)$ is a tensor product of $U \otimes V$ and $W$. Then the statement follows.

To show this by the universal property let $\phi': (U \otimes V) \times W \to P$ be bilinear. Then there is exactly one morphism $\gamma: U \otimes (V \otimes W) \to P$ given by $\gamma(e_i \otimes (f_j \otimes g_k)) = \phi'(e_i \otimes f_j, g_k)$, since $((e_i \otimes (f_j \otimes g_k)))_{1 \leq i \leq n, 1 \leq j \leq m, 1\leq k \leq r}$ is a basis of $U \otimes (V \otimes W) \to P$.

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marked as duplicate by Alex Provost, Lord Shark the Unknown, user296602, hardmath, Community Sep 1 '17 at 7:02

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This is Proposition 12.9 of Lee.

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