Let $M$ and $N$ be $R$-modules with $R$ a commutative ring with identity. When we calculate $Tor\_i^R(M,N)$, usually first we choose projecive resolutions $P\_.$ and $Q\_.$ of $M$ and $N$, then we calculate the $i$-th homology group of the complex $P\_.\otimes Q\_.$. My question is: can we choose an injective resolution $I^.$ of $M$ and a projective resolution $Q\_.$ of $N$, and $Tor\_i^R(M,N)$ is just the $i$-th cohomology group of the cochain complex $I^.\otimes Q\_.$? 


The cochain groups of $I^.\otimes Q\_.$ is defined by N.E.Steenrod as follows: $(I^.\otimes Q\_.)^n=\sum\_{i=0}^{\infty}I^{n+i}\otimes Q\_i$, the coboundary operator is defined by: $d(r\otimes q)=\delta(r)\otimes q+(-1)^{n+i}r\otimes \partial(q)$.



Calculation of $Tor_i^R(M,N)$ using an injective resolution of M and a projective resolution of N.
It is enough to replace one of the objects with its projective resolution. For example, you can replace $N$ with its projective resolution $Q$. Then the cohomology of the complex $M \otimes Q$ is equal to $Tor$'s. On the other hand, after that you can replace $M$ with ANY complex $C$ quasiisomorphic to it, for example with its injective resolution, and the cohomology of $C\otimes Q$ still will be isomorphic to $Tor$'s. The reason for this is the fact that if $C$ is acyclic then $C\otimes Q$ is also acyclic.


So, the answer is yes, you can.


