I'm having trouble with this exercise from Elements of the Representation Theory of Associative Algebras I: Techniques of Representation Theory.


The exercise in question is from chapter IV.


So, let $k$ be an algebraically closed field and let $A$ be a finite-dimensional $k$-algebra, and $M$ a left $A$-module containing no projective direct summands. 


Say $F = \underline{Hom}\_{A^{op}}(M, -) $ and $G = Tor\_1^{A^{op}}(M, -)$. The exercise asks to prove that $F \cong G$.


Now I have been thinking about this exercise for a while, and I have no idea how to prove it.


One thing to note, the functor $F$ is the projective stabilization of the $Hom$ functor, and it takes a module $N$ and sends it to the module $Hom(M, N) / \mathcal{P}(M, N)$, where $\mathcal{P}(M, N)$ is the submodule consisting of all morphisms $f: M \rightarrow N$ that factor through a projective module.


For the life of me, I don't know how to prove this. I have tried to construct a surjective morphism from $Hom(M, N)$ to $Tor(M, N)$ having kernel $\mathcal{P}(M, N)$. This seems like sort of a lot of work to build up $Tor(M, N)$ from scratch and then figuring out where to send a morphism $f$. And then there would still be the problem of showing that for some $g: N \rightarrow N'$ we have a commutative square which I can't quite format online using the following two maps as the top and bottom lines of the square.


$$Hom(M, N) / \mathcal{P}(M, N) \rightarrow Tor(M, N)$$
$$Hom(M, N') / \mathcal{P}(M, N') \rightarrow Tor(M, N')$$


Thanks!



Showing a functorial isomorphism
Here is a proof of the functorial isomorphism $$\underline{\operatorname{Hom}}\_{A^{\rm{op}}}(\operatorname{Tr} M,-) \cong \operatorname{Tor}\_1^{A^{\rm{op}}}(M,-).$$


Since $M$ has no projective summands, we can choose a right $A$-module $X$ without projective summands such that $M \cong \operatorname{Tr} X$ and $X \cong \operatorname{Tr} M$, where $\operatorname{Tr}$ denotes the transpose. For ease of notation let $B=A^{\rm{op}}$.


Since $X$ has no projective summands, the Auslander-Reiten formula gives a functorial isomorphism $$\underline{\operatorname{Hom}}\_B(X,-) \cong D \operatorname{Ext}^1\_B(-,D \operatorname{Tr} X),$$ where $D$ denotes the duality $D=\operatorname{Hom}\_k(-,k)$. For any right $B$-module $U$, there is an isomorphism $$D\operatorname{Tor}\_1^B(U,-) \cong \operatorname{Ext}^1\_B(-,DU).$$ In our case we get $$\underline{\operatorname{Hom}}\_B(X,-) \cong D \operatorname{Ext}^1\_B(-,D \operatorname{Tr} X) \cong \operatorname{Tor}\_1^B(\operatorname{Tr} X,-),$$ which finishes the proof. 


