Hey. I'm working with Bredon's equivariant cohomology. At some point I need to compute the $4$th equivariant cohomology group of $S^1 \x D^3$ relatively to its boundary for the antipodal action of $\mathbb{Z}_2$.

I found a paper that just use "equiviriant poincaré duality" to bring the problem to compute the $0$th equivariant homology which is said to be $\mathbb{Z}_2$. This sounds trivial, but Bredon does not define what equivariant homology is, just cohomology, hence he does not talk of Poincare duality either (At least not in the 30 first pages of his paper "Equivariant Cohomology Theories" I use.).

I'm trying to search for a nice definition of equivariant homology, but every paper I find use concept I don't master well or at all (Vector bundles, groupoids, Borel-Moore Homology). I also tried to design my own definition of equivariant homology, simply letting $H_0$ be the quotient of equivariant chains $H_0 = C_0^{\mathbb Z_2}/\partial(C_1^{\mathbb Z_2})$ with $C_i^{\mathbb Z_2} = \lbrace c \in C_i| (1+\mathbb Z) \curvearrowright c = c \rbrace$. But using the relative exact sequence of chain, one finds $C_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2)$ is a subgroup of $C_0(S^1\times D^3, S^1 \times S^2)$, which is $0$, so $H_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2) \approx 0$, not $\mathbb Z_2$ !

Many thanks for any help !