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I am having trouble understanding a part of a proof given in Dummit and Foote relating to tensor algebras. The theorem states: Suppose $M$ is any $R$ module where $R$ is a commutative ring with $1$. Then $\mathcal{T}(M)$ is an $R$-algebra with multiplication defined by the mapping $(m_1\otimes\cdots\otimes m_i)(m_1'\otimes\cdots\otimes m_j')=m_1\otimes\cdots\otimes m_i\otimes m_1\otimes\cdots\otimes m_j'$ and extended to sums via distributive law. With respect to this multiplication $\mathcal{T}^i(M)\mathcal{T}^j(M)\subseteq\mathcal{T}^{i+j}(M)$.

The proof starts as follows:The map $M\times\cdots\times M\times M\times\cdots \times M\rightarrow\mathcal{T}^{i+j}(M)$ defined by $(m_1,...,m_i,m_1',...,m_j')\mapsto m_1\otimes\cdots\otimes m_i\otimes m_1'\otimes\cdots\otimes m_j'$ is $R$-multilinear, so induces a bilinear map $\mathcal{T}^i(M)\times\mathcal{T}^j(M)$ to $\mathcal{T}^{i+j}(M)$.

Why does the multilinear map above induce this bilinear map? Does this somehow follow from the universal property for tensor products? I know in general multilinear maps induce homomorphisms from tensor products, but that doesn't seem to be what's going on here.

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You have to check that $\varphi \colon T^i(M) \times T^j \rightarrow T^{i+j}(M), \big((m_1 \otimes \ldots \otimes m_i) ,(m^\prime_1 \otimes \ldots \otimes m^\prime_j)\big) \mapsto m_1 \otimes \ldots \otimes m_i \otimes m^\prime_1 \otimes \ldots \otimes m^\prime_j \\$ is bilinear. For this you can use the R-multi lin. of the given map.

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