TITLE: A similar construction to Ext, can we describe it better and does it have any use?
QUESTION [4 upvotes]: Let $R$ be a ring and $\text{Mod}\,R$ the category of $R$ modules. For two $R$-modules $X,Y$ one can define $\text{Ext}_R^n(X,Y)$ as follows. We take an injective resolution $0\rightarrow Y\rightarrow I_0 \rightarrow I_1 \rightarrow \dots,$ throw away $Y$ and apply $\text{Hom}_R(X,-)$ to obtain the cochain complex $0 \rightarrow \text{Hom}_R(X,I_0) \rightarrow \text{Hom}_R(X,I_1) \rightarrow \dots,$ denoted by $C_Y$. Then, we define $\text{Ext}_R^n(X,Y)$ as the $n$'th cohomology of $C_Y$. The horseshoe lemma shows, that a short exact sequence $0\rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ of $R$-modules induces a short exact sequence of cochain complexes $0\rightarrow C_L \rightarrow C_M \rightarrow C_N \rightarrow 0,$ which induces a long exact sequence $0\rightarrow \text{Ext}^0_R(X,L) \rightarrow \text{Ext}^0_R(X,M)\rightarrow \text{Ext}^0_R(X,N)\rightarrow \text{Ext}^1_R(X,L)\rightarrow \text{Ext}^1_R(X,M)\rightarrow \text{Ext}^1_R(X,N) \rightarrow ...,$ where $\text{Ext}^0_R(X,Y) = \text{Hom}_R(X,Y)$.
Now my question is:
What happens, if we do the same with a projective resolution?
That is, we take a projective resolution $\dots \rightarrow P_1\rightarrow P_0 \rightarrow Y \rightarrow 0$ throw away $Y$ and apply $\text{Hom}_R(X,-)$ to obtain the chain complex $\dots \rightarrow \text{Hom}_R(X,P_1) \rightarrow \text{Hom}_R(X,P_0) \rightarrow 0,$ denoted by $D_Y$. Then we define $F^n (Y)$ as the $n$'th homology of $D_Y$. Again the horseshoe lemma shows, that a short exact sequence $0\rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ of $R$-modules induces a short exact sequence of chain complexes $0\rightarrow D_L \rightarrow D_M \rightarrow D_N \rightarrow 0$. In fact, we only need that $\text{Hom}_R(X,-)$ is an additive functor for that. The short exact sequence of chain complexes induces a long exact sequence $\dots \rightarrow F^1(L) \rightarrow F^1(M)\rightarrow F^1(N)\rightarrow F^0(L) \rightarrow F^0(M)\rightarrow F^0(N)\rightarrow 0$. However, this time we have $F^0(Y) \neq \text{Hom}_R(X,Y)$ in general. We had equality before because $\text{Hom}_R(X,-)$ is left-exact but this time we would need right-exactness. Nevertheless, one can show that this construction induces a natural transformation $F^0 \rightarrow \text{Hom}_R(X,-)$.
Now my question is: Can we describe $F^n$ better? In particular for $n=0$? Is this contrsuction useful in any way?

REPLY [8 votes]: Suppose for now that $X$ is finitely presented. Set $X^\vee = \operatorname{Hom}_R(X,R)$, so that there is a natural transformation  $X^\vee\otimes_R M\to \operatorname{Hom}_R(X,M)$ which is an isomorphism for $M$ a projective $R$-module: For finite rank free modules, this is immediate, for general free modules this follows since both sides commute with filtered colimits (this uses the finite presentation hypothesis), and for general projective modules this follows since they are retracts of free modules. Thus the complex $D_Y(X)$ is isomorphic to the complex $...\to X^\vee\otimes_R P_1\to X^\vee\otimes_R P_0\to 0$, so that its homology is given by $F^n(X) = \operatorname{Tor}^n_R(X^\vee,Y)$.
A general $R$-module $X$ is the filtered colimit of the diagram of maps from a finitely presented module $f:X'\to X$. Pulling out the colimit gives an isomorphism of chain complexes $D_Y(X)\cong\varprojlim_{f:X'\to X,X'\text{ f.p.}} D_Y(X')$. There is a map $H_n(D_Y(X))\to \varprojlim_{f:X'\to X,X'\text{ f.p.}} F^n(X')$; however, as Denis-Charles Cisinski mentions filtered limits are not exact, so that this map does not have to be an isomorphism. The map to $\operatorname{Hom}_R(X,Y)$ factors through it via the limit of the maps $(X')^\vee\otimes_R Y\to\operatorname{Hom}(X',Y)$.
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