Equivalent condition for $\mathrm{Tor}_{n+1}^R(G,N)=0$ I am reading Relative Homological Algebra by Enochs. In the proof of lemma 9.1.4, he said '$0 \to S \otimes N \to P_n \otimes N$ is exact since $\mathrm{Tor}_{n+1}(G,N)=0$'. Besides, he said '$0 \to S \otimes M \to P_n \otimes M$ is exact and so $\mathrm{Tor}_{n+1}(G,M)=0$' (see the following picture). How can I prove this? I know
$$
\mathrm{Tor}_{n+1}(G,M)=0 \Leftrightarrow \ker(d^{n+1}\otimes 1_M)=\mathrm{Im}(d^{n+2}\otimes 1_M),
$$
but I am stuck here. I think $S$ means $\ker(d^n)$ here, the so-called partial projective resolution is not a projective resolution. Thanks in advance

 A: $\newcommand{\Tor}{\operatorname{Tor}}$If you pick a partial projective resolution of $G$ of the form
$$0\to K_n\to P_n \to\cdots \to P_1\to P_0\to G$$
and consider only the portion $0\to K_n\to P_n \to K_{n-1}\to 0$ (where $K_{n-1}$ is just the image of the arrow $P_n\to P_{n-1}$ then the long exact sequence for this looks like
$$0\to \Tor_1(K_{n-1},N) \to S\otimes N\to P_n \otimes N\to K_{n-1}\otimes N\to 0.$$
Now, you can show by induction (this is called ''dimension shifting'') that if you write $K_n$ for the kernel of the map $P_n\to P_{n-1}$, then $\Tor_p(G,K_n) \cong \Tor_{p+n+1}(G,N)$.
Indeed, the LES for $-\otimes N$ applied to $$0\longrightarrow K_n \to P_n\longrightarrow K_{n-1}\longrightarrow 0$$
(since $K_{n-1}$ is also the image of the map $P_n\to P_{n-1}$)
gives you isomorphisms
$$\Tor_{p+1}(K_{n-1},N)\longrightarrow \Tor_p(K_n,N).$$
Applying this many times gives you the isomorphism
$$\Tor_{n+1}(K_{-1},N) \longrightarrow 
 \Tor_{n}(K_0 ,N) \longrightarrow  \Tor_{n-1}(K_1,N)\longrightarrow$$
$$\cdots\longrightarrow 
\Tor_2(K_{n-2},N) \longrightarrow\Tor_1(K_{n-1},N)$$
which is what you wanted (notice $K_{-1}=G$).
