commutativity of torsion functor For a ring $R$ and finitely generated $R$ modules $U,W$,
$${\rm Tor}_i(U,W)={\rm Tor}_i(W,U)$$ for all $i$.
I saw proof in Hatcher's book, but I can understand that proof.
may be I see another proof?
 A: The proof depends on the definitions of $\newcommand{\Tor}{\textrm{Tor}}$ $\Tor^R_\bullet(U, W)$ used, so I'll follow Hatcher's definition using free resolutions of $U$. This admits a reasonably direct proof not invoking much abstract nonsense.
At first, let $R$ be (not necessarily commutative) ring. For each right $R$-module $U$, fix a free resolution $F_\bullet \twoheadrightarrow U$, and for each left $R$-module $W$, fix a flat resolution $M_\bullet \twoheadrightarrow W$. Let us write $Z_n$ for the kernel of $M_n \to M_{n-1}$, so that we get short exact sequences
$$0 \longrightarrow Z_0 \longrightarrow M_0 \longrightarrow W \longrightarrow 0$$
$$0 \longrightarrow Z_{n+1} \longrightarrow M_{n+1} \longrightarrow Z_n \longrightarrow 0$$
for each natural number $n$. Because each $F_m$ is free, the tensored sequences
$$0 \longrightarrow F_m \otimes_R Z_0 \longrightarrow F_m \otimes_R M_0 \longrightarrow F_m \otimes_R W \longrightarrow 0$$
$$0 \longrightarrow F_m \otimes_R Z_{n+1} \longrightarrow F_m \otimes_R M_{n+1} \longrightarrow F_m \otimes_R Z_n \longrightarrow 0$$
are also exact, and therefore we have short exact sequences of chain complexes. The zig-zag lemma of homological algebra then implies we have these long exact sequences:
$$\cdots \longrightarrow \Tor^R_{i+1}(U, M_0) \longrightarrow \Tor^R_{i+1}(U, W) \longrightarrow \Tor^R_i(U, Z_0) \longrightarrow \Tor^R_i(U, M_0) \longrightarrow \cdots$$
$$\cdots \longrightarrow \Tor^R_{i+1}(U, M_{n+1}) \longrightarrow \Tor^R_{i+1}(U, Z_n) \longrightarrow \Tor^R_i(U, Z_{n+1}) \longrightarrow \Tor^R_i(U, M_{n+1}) \longrightarrow \cdots$$
Now, each $M_n$ is flat, so the sequence
$$\cdots \longrightarrow F_2 \otimes_R M_n \longrightarrow F_1 \otimes_R M_n \longrightarrow F_0 \otimes_R M_n \longrightarrow U \otimes_R M_n \longrightarrow 0$$
remains exact, and therefore $\Tor^R_{i+1}(U, M_n) = 0$ for all natural numbers $i$. Hence, there are isomorphisms
$$\Tor^R_1(U, Z_n) \cong \Tor^R_2(U, Z_{n-1}) \cong \cdots \cong \Tor^R_{n+1}(U, Z_0) \cong \Tor^R_{n+2}(U, W)$$
and exact sequences
$$0 \longrightarrow \Tor^R_1(U, W) \longrightarrow U \otimes_R Z_0 \longrightarrow U \otimes_R M_0 \longrightarrow U \otimes_R W \longrightarrow 0$$
$$0 \longrightarrow \Tor^R_1(U, Z_n) \longrightarrow U \otimes_R Z_{n+1} \longrightarrow U \otimes_R M_{n+1} \longrightarrow U \otimes_R Z_n \longrightarrow 0$$
It is easy to check that 
$$H_0(U \otimes_R M_\bullet) \cong U \otimes_R W$$
and mucking around with kernels and cokernels one finds that
$$H_1(U \otimes_R M_\bullet) \cong \Tor^R_1(U, W)$$
$$H_{n+2}(U \otimes M_\bullet) \cong \Tor^R_1(U, Z_n) \cong \Tor^R_{n+2}(U, W)$$
and thus we have proven the following:
Proposition. $\Tor^R_\bullet(U, W)$ can be computed using


*

*either a free resolution of $U$,

*or a flat resolution of $W$. 　　◼


Now, let us assume $R$ is commutative. Assume $F_\bullet \twoheadrightarrow U$ is a free resolution. Commutativity of $R$ implies that the two chain complexes
$$\cdots \longrightarrow F_2 \otimes_R W \longrightarrow F_1 \otimes_R W \longrightarrow F_0 \otimes_R W$$
$$\cdots \longrightarrow W \otimes_R F_2 \longrightarrow W \otimes_R F_1 \longrightarrow W \otimes_R F_1$$
are isomorphic. But the homology of the first one computes $\Tor^R_\bullet (U, W)$ and the homology of the second one computes $\Tor^R_\bullet (W, U)$, we have
$$\Tor^R_\bullet (U, W) \cong \Tor^R_\bullet (W, U)$$
as required.
