# universal property of the $n$-fold tensor product

How can one prove the universal property of the $n$-fold tensor product from the binary universal property, plus associativity?

Specifically, I am thinking of the situation in Dummit and Foote $\S$10.4, where after proving Corollary 12 (the correspondence between $R$-bilinear maps $\phi: M \times N$ and $R$-module homomorphisms $\Phi: M \otimes_R N$), and exhibiting the natural isomorphism between $(M \otimes N) \otimes L$ and $M \otimes (N\otimes L)$, the authors state the correspondence between $n$-multilinear maps $\phi: M_1 \times \dots \times M_n$ and $R$-module homomorphisms out of the (now unambiguously defined) $M_1 \otimes \dots \otimes M_n$. The authors say this correspondence follows from the binary one, but I don't see how to show it.

More generally, can we define a tensor product $\bigotimes_{i \in I}M_i$ of some arbitrarily indexed modules $M_i$, and prove a universal property for it?

## 1 Answer

$\def\id{{\rm id}} \def\phi{\varphi} \def\motimes{\otimes \cdots \otimes} \def\mtimes{\times \cdots \times}$

I won't address your second question about infinite tensor products, but see here.

For the phenomenon of deriving general associativity from associativity, the comments of dari grinberg, in this post, are very good.

The proof of the universal property of the $n$-fold tensor product by induction follows the same pattern as the proof of Theorem 14 in Section 10.4 in Dummit and Foote.

Take as the model of the $n$-fold tensor product the left bracketed version, e.g for $n = 4$, $(((A\otimes B)\otimes C)\otimes D)$. Due to this specific construction the sequence of tensor products $M_1, M_1 \otimes M_2, M_1 \otimes M_2 \otimes M_3, \dots$ comes with canonical bilinear maps $$i_n : (M_1 \otimes \cdots \otimes M_{n-1}) \times M_n \to M_1 \otimes \cdots \otimes M_{n}.$$
Define $$j_n : M_1 \times \cdots \times M_{n} \to M_1 \otimes \cdots \otimes M_{n}$$ recursively by $j_1 = \id_{M_1}$ and $$j_n = i_{n} \circ (j_{n-1} \times \id_{M_n}).$$ Then $j_n$ is multilinear. What we want to show is that for any multilinear map $$\phi: M_1 \times \cdots \times M_{n} \to N,$$ there exists a unique linear map $$\tilde \phi: M_1 \otimes \cdots \otimes M_{n} \to N,$$ such that $\phi = \tilde\phi \circ j_n$.

For fixed $x \in M_n$, define $\phi_x: M_1 \times \cdots \times M_{n-1} \to N$ by $$\phi_x(m_1, \dots, m_{n-1}) = \phi(m_1, \dots, m_{n-1}, x).$$ By the appropriate induction hypothesis, there exists a unique linear $$\tilde \phi_x : M_1 \otimes \cdots \otimes M_{n-1} \to N$$ such that $\tilde \phi_x \circ j_{n-1} = \phi_x$. Define $$\phi': (M_1 \motimes M_{n-1}) \times M_n \to N$$ by $\phi'(t, x) = \tilde \phi_x(t)$. Then $\phi'$ is bilinear and satisfies \begin{align*} \phi'(m_1 \motimes m_{n-1}, x) &=\tilde \phi_x(m_1 \motimes m_{n-1}) \\ &= \phi_x(m_1, \cdots, m_{n-1}) \\&= \phi(m_1, \dots, m_{n-1}, x). \end{align*} Otherwise expressed, this is $\phi' \circ(j_{n-1} \times \id_{M_n}) = \phi$. Next we need a uniqueness statement for $\phi'$. Suppose $\psi'$ is another bilinear map $$\psi': (M_1 \motimes M_{n-1}) \times M_n \to N$$ such that $\psi'(m_1 \motimes m_{n-1}, x) = \phi(m_1, \dots, m_{n-1}, x)$. Use the uniqueness of $\tilde \phi_x$ for each $x$ to show that $\psi' = \phi'$.

Since $\phi'$ is bilinear, there exists a unique linear $\phi: M_1 \motimes M_n \to N$ such that $\tilde \phi\circ i_n = \phi'$. Now we have $$\tilde \phi \circ j_n = \tilde \phi \circ i_n \circ( j_{n-1} \times \id_{M_{n-1}}) = \phi' \circ ( j_{n-1} \times \id_{M_{n-1}}) = \phi.$$ Finally, use the two uniqueness statements obtained along the way to show that $\tilde \phi$ is the unique linear map with this property.

• Great! I don't see how to combine the uniqueness statements at the end, though... Commented Feb 8, 2018 at 4:49