# Product of elements in tensor algebra

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.

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.