A Universe of Sorts

Let $A$ be a commutative ring, $P$ an $A$ -module. The functors $Tor_i^A(-, P)$ are defined in such a way that

$Tor_0^A(-,P) = - \otimes_A P$
For any short exact sequence of $A$ -modules $0 \to L \to M \to N \to 0$ , you get a long exact sequence.

\dots \to Tor_{n+1}^A(L,P) \to Tor_{n+1}^A(M,P) \to Tor_{n+1}^A(N,P) \to Tor_n^A(L,P) \to Tor_n^A(M,P) \to Tor_n^A(N,P) \to \dots

which, on the right side, stops at

\dots \to Tor_1^A(L,P) \to Tor_1^A(M,P) \to Tor_1^A(N,P) \to L \otimes_A P \to M \otimes_A P \to N \otimes_A P \to 0

23:44  isekaijin can you describe the existence proof of Tor? :)23:45  A projective resolution is a chain complex of projective A-modules “... -> P_{n+1} -> P_n -> ... -> P_1 -> P_0 -> 0” that is chain-homotopic to “0 -> P -> 0”.23:45  And you need the axiom of choice to show that it exists in general.23:45  Now, projective A-modules behave much more nicely w.r.t. the tensor product than arbitrary A-modules.23:46  In particular, projective modules are flat, so tensoring with a projective module *is* exact.23:47  So to compute Tor_i(M,P), you tensor M with the projective resolution, and then take its homology.23:47  To show that this is well-defined, you need to show that Tor_i(M,P) does not depend on the chosen projective resolution of P.23:48  bollu: just use the axiom of choice like everyone else23:48  why do you need to take homology?23:48  That's just the definition of Tor.23:49  Okay, to show that Tor does not depend on the chosen projective resolution, you use the fact that any two chain-homotopic chains have the same homology.23:49  right23:49  Which is a nice cute exercise in homological algebra that I am too busy to do right now.23:49  whose proof I have seen in hatcher23:49  :)23:49  Oh, great.23:49  thanks, the big picture is really useful

§ The Tor Functor