Theorem: If $(T_1, \iota_1)$ and $(T_2, \iota_2)$ both satisfy the universal property for the tensor product of $V$ and $W$, then there exists a unique isomorphism $\psi: T_1 \to T_2$ making the diagram commute.

Proof:

Step 1: Apply universal property of $(T_1, \iota_1)$ to $(T_2, \iota_2)$.

Since $(T_2, \iota_2)$ satisfies the universal property, the bilinear map $\iota_2: V \times W \to T_2$ induces a unique linear map $\psi: T_1 \to T_2$ such that: $\psi \circ \iota_1 = \iota_2$

Step 2: Apply universal property of $(T_2, \iota_2)$ to $(T_1, \iota_1)$.

Similarly, since $(T_1, \iota_1)$ satisfies the universal property, the bilinear map $\iota_1: V \times W \to T_1$ induces a unique linear map $\phi: T_2 \to T_1$ such that: $\phi \circ \iota_2 = \iota_1$

Step 3: Compose to show $\phi \circ \psi = \text{id}_{T_1}$.

Consider the composition $\phi \circ \psi: T_1 \to T_1$. We have: $(\phi \circ \psi) \circ \iota_1 = \phi \circ (\psi \circ \iota_1) = \phi \circ \iota_2 = \iota_1$

But the identity map $\text{id}_{T_1}: T_1 \to T_1$ also satisfies: $\text{id}_{T_1} \circ \iota_1 = \iota_1$

By uniqueness in the universal property of $(T_1, \iota_1)$ applied to the bilinear map $\iota_1$ itself, we must have: $\phi \circ \psi = \text{id}_{T_1}$

Step 4: Similarly show $\psi \circ \phi = \text{id}_{T_2}$.

Consider $\psi \circ \phi: T_2 \to T_2$: $(\psi \circ \phi) \circ \iota_2 = \psi \circ (\phi \circ \iota_2) = \psi \circ \iota_1 = \iota_2$

Also: $\text{id}_{T_2} \circ \iota_2 = \iota_2$

By uniqueness in the universal property of $(T_2, \iota_2)$: $\psi \circ \phi = \text{id}_{T_2}$

Step 5: Conclude $\psi$ is an isomorphism.

Since $\psi: T_1 \to T_2$ has two-sided inverse $\phi$, it is an isomorphism. Moreover, $\psi$ is the unique map with $\psi \circ \iota_1 = \iota_2$ by the universal property.

Conclusion: The tensor product is unique up to unique isomorphism. Any two constructions satisfying the universal property are canonically isomorphic. ∎