Cohomology of fixed point subspaces Suppose $M$ is a smooth manifold and $\phi : M \to M$ is a homeomorphism whose fixed point set is a smooth submanifold $M_{\phi}$.  Is there any relation between the cohomology ring of $M_{\phi}$ and the cohomology ring of $M$ augmented by its natural $\mathbf{Z}[\phi]$-algebra structure?  Is it too much to try to compute the former in terms of the latter?  I am sorry for the vagueness of this; I am not sure precisely what question I want to be asking, but I am posting this to find out if I am completely wasting my time by looking for such interactions.
 A: Suppose $G$ is a finite group on a space $X$. Then [Swan, Richard G. A new method in fixed point theory. Bull. Amer. Math. Soc. 65 1959 128--130. MR0107238 (21 #5963); Swan, Richard G. A new method in fixed point theory. Comment. Math. Helv. 34 1960 1--16. MR0115176 (22 #5978)] constructs a spectral sequence which has $$E_2^{i,j}=\hat H^i(G,H^j(X))$$ converging to a certain filtered graded group $J^\bullet(X)$. The spectral sequence degenerates at $E_2$ if the action of the group acts trivially. Here cohomology of spaces is Cech cohomology, with any coefficients you like, and $\hat H$ is Tate cohomology of groups.
Now, suppose $G$ is cyclic, that $X=M$ is a manifold, and that each point in $M$ with non-trivial stabilizer is actually fixed by the whole of $G$. Then Swan shows that $J^\bullet(X)$ is isomorphic to $J^\bullet(X^G)$.
Over a field, an under the above conditions, this gives information about $H^\bullet(X^G)$ from information on $H^\bullet(X)$ and its $G$-module structure.
Swan constructs a multiplicative structure on the spectral sequence, and this gives information on the cohomology ring of the invariants.
A: Example: Let $N\subset\mathbb R^n$ be a compact submanifold. Choose a diffeomorphism $h\colon\mathbb R\to\mathbb R$ satisfying $h(t)\geq t$ for all $t$, and $h(t)=t$ iff $t=0$. Let $M=\mathbb R^n\times\mathbb R$ and $\phi\colon M\to M$, $(x,t)\mapsto(x,h(t)+d(x,N))$. The fixed point set of $\phi$ is $N\times0$, but the homotopy class of $\phi$ does not depend on $N$. (You can easily modify this example so that $M$ is compact.)
