Positive Tor vanishes when the resolved variable is projective

Put $K_{-1}=M$, let $q_0:P_0\to K_{-1}$ be the augmentation, and put $K_0=\ker q_0$. For $j\ge 1$, exactness factors the differential as $d_j={\iota }_{j-1}{q}_j$, where $q_j:P_j\twoheadrightarrow K_{j-1}$ is surjective, $K_j=\ker q_j$, and ${\iota }_j:K_j\hookrightarrow P_j$ is inclusion.

If $K_{j-1}$ is projective, [F1] applied to its identity gives a section $s_j:K_{j-1}\to P_j$ of $q_j$. The map $p\mapsto p-s_j{q}_j(p)$ takes values in $K_j$; write it $r_j:P_j\to K_j$. Then $r_j{\iota }_j=1$ and ${\iota }_j{r}_j+s_j{q}_j=1$. A retract of a projective module is projective: for $f:K_j\to B$ and a surjection $E\to B$, lift $f{r}_j:P_j\to B$ and restrict the lift along ${\iota }_j$. Thus $K_j$ is projective.

Starting with the projective $K_{-1}=M$, repeat step 2.1 through $j=i+1$. This is a finite succession of existential choices for the fixed integer $i$, not a simultaneous choice of splittings in all degrees. It needs neither AC nor dependent choice.

Write $T(f)=1_N\otimes f$. If $z\in N{\otimes }_R{P}_i$ is a cycle, then $T({\iota }_{i-1})T(q_i)z=0$. Since $T(r_{i-1})T({\iota }_{i-1})=1$, it follows that $T(q_i)z=0$. The identity from step 2.1 therefore gives $z=T({\iota }_i)T(r_i)z$.

Set $w=T(s_{i+1})T(r_i)z\in N{\otimes }_R{P}_{i+1}$. Then $T(d_{i+1})w=T({\iota }_i)T(q_{i+1})T(s_{i+1})T(r_i)z=T({\iota }_i)T(r_i)z=z$. Every cycle is a boundary, so $H_i(N{\otimes }_R{P}_{\bullet })=0$.

For a projective right module $N$ and any supplied right projective resolution $Q_{\bullet }\to N$, perform the same construction with right-linear maps and apply $T(f)=f\otimes 1_M$. The identities in steps 4.1 and 5.1 are unchanged and give $H_i(Q_{\bullet }{\otimes }_{R}M)=0$. Since $i>0$ was arbitrary, [F2] gives both claimed vanishings for every supplied resolution, without selecting a family of splittings over all $i$.