# Abstract Nonsense

## Adjoint Isomorphism for Tensor and Hom (Pt. II)

Point of Post: This is a continuation of this post.

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From all of this we can conclude that $\eta$ is a well-defined map (yay!). Let’s now verify that it is a group isomorphism. To see that $\eta$ is a homomorphism we merely note that

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\begin{aligned}\eta(f+f')(m)(n) &=(f+f')(m\otimes n)\\ &=f(m\otimes n)+f'(m\otimes n)\\ &=\eta(f)(m)(n)+\eta(f')(m)(n)\\ &=(\eta(f)+\eta(f'))(m)(n)\end{aligned}

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and since $m,n$ were arbitrary the desired equality holds. Lastly, let’s check that $\eta$ is a bijection. Injectivity is simple enough, if $\eta(f)=\eta(g)$ then we see that $f$ and $g$ agree on all simple tensors in $M\otimes_R N$ and so (because the simple tensors are a generating set) must be equal on all of $M\otimes_R N$. For surjectivity, let $F:M\to\text{Hom}_S(N,L)$ be an arbitrary $(A,R)$-map. Define the map $f:M\times N\to L$ given by $f(m,n)=F(m)(n)$. We claim that $f$ is an $R$-biadditive $(A,S)$-bimap. Indeed:

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\begin{aligned}f(amr,ns) &= F(amr)(ns)\\ &=F(amr)(n)s\\ &(aF(m)r)(n)s\\ &=aF(m))(rn)s\\ &=af(m,rn)s\end{aligned}

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Thus, by the universal properties of the tensor product $f$ lifts to an $(A,S)$-map $\widetilde{f}:M\otimes_R N\to L$ such that $\widetilde{f}(m\otimes n)=F(m)(n)$. Note then that $\eta(\widetilde{f})(m)(n)=\widetilde{f}(m\otimes n)=F(m)(n)$ for all $m,n$ and so $\eta(\widetilde{f})=F$. Thus, surjectivity, and thus bijectivity, and thus isomorphism follow.

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Let us fix $L,N$ and show that in $M$ $\eta_{(M,N,L)}$ is a natural transformation. Indeed, suppose that $f:M\to M'$ is an $(A,R)$-map then we wish to show the commutativity of the following diagram

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$\begin{matrix}\text{Hom}_{(A,S)}(M'\otimes_R N,L) & \overset{\eta_{(M',N,L)}}{\longrightarrow} & \text{Hom}_{(A,R)}(M',\text{Hom}_S(N,L))\\ ^{(f\otimes1_N)^\ast}\big\downarrow & & \big\downarrow ^{f^\ast}\\ \text{Hom}_{(A,S)}(M\otimes_R N,L) & \underset{\eta_{(M,N,L)}}{\longrightarrow} & \text{Hom}_{(A,R)}(M,\text{Hom}_S(M,L))\end{matrix}$

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but this is easy:

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$f^\ast((\eta_{(M',N,L)})(g))(m)(n)=(\eta_{(M',N,L)}(g)(f(m)))(n)=g(f(m)\otimes n)$

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and

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\begin{aligned}(\eta_{(M,N,L)}(f\otimes 1_N)^\ast(g))(m)(n) &=(\eta_{(M,N,L)}(g\circ(f\otimes 1_N))(m)(n)\\ &=(g\circ (f\otimes 1_N))(m\otimes n)\\ &=g(f(m)\otimes n)\end{aligned}

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The case for $\nu$ is basically the same as the case for $\eta$, and so we admit it. $\blacksquare$

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Ok, so getting a little more down to earth, the above basically says that if we take a commutative ring $R$ and and take three $R$-modules (thought of, in the usual way, as $(R,R)$-bimodules) and noting that $\eta$ and $\nu$ are both $R$-maps in this context, that’s pretty easy to see

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$\eta(rf)(m)(n)=(rf)(m\otimes n)=rf(m\otimes n)=r\eta(f)(m)(n)$

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for example, we have the following corollary (which is the case we’ll most often use):

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Theorem: Let $R$ be a commutative unital ring and $M,N,L$ be unital $R$-modules. Then, the above maps $\nu$ and $\eta$ give $R$-isomorphisms

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\begin{aligned}&\text{Hom}_R(M\otimes_R N,L)\cong\text{Hom}_R(M,\text{Hom}_R(N,L))\\ &\text{Hom}_R(M\otimes_R N,L)\cong\text{Hom}_R(N,\text{Hom}_R(M,L))\end{aligned}

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.