Prove $\phi$ to be isomorphism (an exercise in commutative algebra)

Question

The question is from my course exercise of commutative algebra and I am asking here as I was unable to make any significant progress in it. $\DeclareMathOperator{\Hom}{Hom}$

Suppose there exists $f \in \Hom(M, \Hom(N,P))$ , then $f$ defines an $A$-bilinear map from $M\times N \to P$ given by $f'(x,y)=f(x)(y)$.

Define $\phi : \Hom(M, \Hom(N,P)) \to \Hom(M \otimes_A N, P)$ by $\phi : \phi (f) (m \otimes n) =f(m)(n)$. So, using definition of $\phi$ we can see that $\phi (f) = g \in \Hom(M \otimes_A N, P)$ where $g$ is the hom $g(m \otimes n)= f(m)(n)$.

Prove that $\phi$ is an isomorphism.

I have proved it is 1-1 but am unable to prove it is onto (I need to prove that for all $f \in \Hom(M \otimes_A N , P)$, there exists an $f \in \Hom( M , \Hom(N,P))$. I am unable to see why this must hold true.

Answer 1

Question: "I have proved it is 1-1 but am unable to prove it is onto.... Can you please elaborate? It will be really helpful to me – Avenger"

Answer: There is an inverse map defined as follows:$\DeclareMathOperator{\Hom}{Hom}$

Let $f: M \rightarrow \Hom_A(N,P)$. This means for every $m\in M$ you get an $A$-linear map

$$f_m: N \rightarrow P.$$

Define the map $\psi: \Hom_A(M,\Hom_A(N,P)) \rightarrow \Hom_A(M\otimes_A N ,P)$

by

$$\psi(f): M\otimes_A N \rightarrow P$$

where

$$\psi(f)(m\otimes n):=f_m(n) \in P.$$

In the other direction there you may do as follows:

Assume $f: M\otimes_A N \rightarrow P$ is $A$-linear. For any element $m\in M$ there is a map

$$ N \rightarrow^{g_m} M\otimes_A N \rightarrow^f P$$

where $g_m(n):=m\otimes n$. This defines a map in the other direction:

$$\phi(f)(m)(n):=f(m\otimes n)$$

and

$$\phi: Hom_A(M\otimes_A N, P) \rightarrow Hom_A(M,Hom_A(N,P)).$$

You should verify that $\phi \circ \psi=\psi \circ \phi=Id$.

Usually in commutative algebra/homological algebra when proving two $A$-modules (or rings etc.) are isomorphic in complete generality, you usually are able to exhibit an inverse map. Completely general statements in homological algebra/catgory theory are seldom "deep". Their proofs are usually short and obvious.

Here are "similar examples" where I construct two maps - one in each direction.

1: To prove that the "magic diagram" is cartesian and

2: An isomorphism of basic open subschemes:

The "magic diagram" is cartesian

Form of basic open set of affine scheme: The intersection of two basic open sets.

Note: The "magic diagram" post is voted down, but if you cannot see it and are interested in the proof, write a new post and I'll post the answer once more.