Universal Multilinear Object (10.4 Corollary 16)

Dummit and Foote Abstract Algebra, section 10.4, corollary 16:

Show that if $R$ is a commutative ring, $A_i$ is an $R$-module, $L$ is an abelian group, and $φ : A_1 \times ... \times A_n → L$ is a multilinear map, then there exists a unique homomorphism of $R$-modules $\Phi : A_1 \otimes ... \otimes A_n → L$ such that $φ = \Phi \circ \iota$, where $\iota$ is the natural map from $A_1 \times ... \times A_n$ to $A_n \otimes ... \otimes A_n$. Conversely, show that if there is a homomorphism of $R$-modules $\Phi : A_1 \otimes ... \otimes A_n → L$ then $φ = \Phi \circ \iota$ is a multilinear map.

Proof: Lemma 1: We have the natural map on simple tensors$$\iota ' : A_1 \otimes ... \otimes A_{n-1} \times A_n → (A_1 \otimes ... \otimes A_{n-1}) \otimes A_n ≅ A_1 \otimes ... \otimes A_n$$ is well defined, and together with the evidently well-defined natural map$$\iota^* : A_1 \times ... \times A_n → A_1 \otimes ... \otimes A_{n-1} \times A_n$$we have the relation $\iota = \iota ' \circ \iota^*$. Proof: The lemma will follow as soon as we show $\iota '$ is well defined. If $(a_1 \otimes ... \otimes a_{n-1}, a_n)=(a_1' \otimes ... \otimes a_{n-1}', a_n')$ then by using these two given equivalencies in the coordinates we clearly have$$\iota '(a_1 \otimes ... \otimes a_{n-1}, a_n)=(a_1 \otimes ... \otimes a_{n-1}) \otimes a_n =$$$$(a_1' \otimes ... \otimes a_{n-1}') \otimes a_n' = \iota '(a_1' \otimes ... \otimes a_{n-1}', a_n')~\square$$Proceed by induction on $n$. For each $a_n∈A_n$ define$$φ_{a_n} : A_1 \times ... \times A_{n-1} → L$$$$φ_{a_n}(a_1,...,a_{n-1})=φ(a_1,...,a_n)$$Note that $φ_{a_n}$ is multilinear$$φ_{a_n}(a_1,...,ra_m+r'a_m',...,a_{n-1})=φ(a_1,...,ra_m+r'a_m',...,a_n)=$$$$rφ(a_1,...,a_m,...,a_n)+r'φ(a_1,...,a_m',...,a_n)=$$$$rφ_{a_n}(a_1,...,a_m,...,a_{n-1})+r`φ_{a_n}(a_1,...,a_m',...,a_{n-1})$$so that by induction there exists a unique homomorphism of $R$-modules$$\Phi_{a_n} : A_1 \otimes ... \otimes A_{n-1} → L$$such that $φ_{a_n}=\Phi_{a_n} \circ \iota^\Delta$ where $\iota^\Delta$ is the natural map from $A_1 \times ... \times A_{n-1}$ to $A_1 \otimes ... \otimes A_{n-1}$. Now define$$σ' : A_1 \otimes ... \otimes A_{n-1} \times A_n → A_1 \otimes ... \otimes A_{n-1}$$$$σ'(a_1 \otimes ... \otimes a_{n-1}, a_n)=a_1 \otimes ... \otimes a_{n-1}$$$$ψ : A_1 \otimes ... \otimes A_{n-1} \times A_n → L$$$$ψ = \Phi_{a_n} \circ σ'$$(Where $a_n$ in the definition of $ψ$ depends on the input) Prove $ψ$ is bilinear by$$ψ(r \sum a_{i,1} \otimes ... \otimes a_{i,n-1} + r' \sum a_{i,1}' \otimes ... \otimes a_{i,n-1}',a_n)=$$$$\Phi_{a_n}(r \sum a_{i,1} \otimes ... \otimes a_{i,n-1} + r' \sum a_{i,1}' \otimes ... \otimes a_{i,n-1}')=$$$$r\Phi_{a_n}(\sum a_{i,1} \otimes ... \otimes a_{i,n-1})+r'\Phi_{a_n}(\sum a_{i,1}' \otimes ... \otimes a_{i,n-1}')=$$$$rψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)+r'ψ(\sum a_{i,1}' \otimes ... \otimes a_{i,n-1}',a_n)$$as well as$$ψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},ra_n+r'a_n')=\Phi_{ra_n+r'a_n'}(\sum a_{i,1} \otimes ... \otimes a_{i,n-1})=$$$$\sum \Phi_{ra_n+r'a_n'}(a_{i,1} \otimes ... \otimes a_{i,n-1})=\sum (\Phi_{ra_n+r'a_n'} \circ \iota^\Delta)(a_{i,1},...,a_{i,n-1})=$$$$\sum φ_{ra_n+r'a_n'}(a_{i,1},...,a_{i,n-1})=\sum φ(a_{i,1},...,a_{i,n-1},ra_n+r'a_n')=$$$$\sum [rφ(a_{i,1},...,a_n)+r'φ(a_{i,1},...,a_n')]=$$$$\sum rφ(a_{i,1},...,a_n) + \sum r'φ(a_{i,1},...,a_n')=$$$$r\sum φ(a_{i,1},...,a_n) + r'\sum φ(a_{i,1},...,a_n')=$$$$rψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)+r'ψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n')$$so that there exists a unique homomorphism of $R$-modules$$\Phi : A_1 \otimes ... \otimes A_n → L$$such that, in part, $ψ = \Phi \circ \iota '$ on simple tensors where $\iota '$ is described in the lemma. Now we can show$$\Phi \circ \iota = \Phi \circ \iota ' \circ \iota^* = ψ \circ \iota^* = \Phi_{a_n} \circ σ' \circ \iota^*$$Since clearly $σ' \circ \iota^* = \iota^\Delta \circ σ$ where$$σ : A_1 \times ... \times A_n → A_1 \times ... \times A_{n-1}$$$$σ(a_1,...,a_n)=(a_1,...,a_{n-1})$$we can continue$$\Phi_{a_n} \circ σ' \circ \iota^* = \Phi_{a_n} \circ \iota^\Delta \circ σ = φ_{a_n} \circ σ = φ$$This completed $⇒$ argument in reverse also shows $ψ = \Phi \circ \iota ' ⇔ φ = \Phi \circ \iota$ since $\iota^*$ is surjective on simple tensors, which is seen to extend to $\Leftarrow$ by$$ψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)=\sum ψ(a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)=$$$$\sum (\Phi \circ \iota ')(a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)= \sum \Phi(a_{i,1} \otimes ... \otimes a_n)=$$$$\Phi(\sum a_{i,1} \otimes ... \otimes a_n) = \Phi((\sum a_{i,1} \otimes ... \otimes a_{i,n-1}) \otimes a_n)=$$$$(\Phi \circ \iota ')(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)$$This establishes the first part. For the converse, observe$$(\Phi \circ \iota)(a_1,...,ra_m+r'a_m',...,a_n)=\Phi(a_1 \otimes ... \otimes (ra_m + r'a_m') \otimes ... \otimes a_n)=$$$$r\Phi(a_1 \otimes ... \otimes a_m \otimes ... \otimes a_n) + r'\Phi(a_1 \otimes ... \otimes a_m' \otimes ... \otimes a_n) =$$$$r(\Phi \circ \iota)(a_1,...,a_m,...,a_n)+r'(\Phi \circ \iota)(a_1,...,a_m',...,a_n)~\square$$