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 Apr14 awarded Yearling Apr14 awarded Yearling Apr14 awarded Yearling Apr11 awarded Nice Question Mar13 comment Deck Transformations of Branched Covers @Ryan: but the homotopy theory of such branched covers is very different from the usual theory, right? For example there is no uniqueness in path lifting and hence there should be no connection between such covers and subgroups of the fundamental group?! Eg there are branched covers over simply connected spaces, I believe complex geometers like to look at branched double covers of a K3 Kummer surface. Do you know something about the deck transformations of a branched cover (for unramified covers given in terms of quotients of the fundamental group?) Mar4 comment Help showing homeomorphism well, it indeed is strongly related to what you want to show. you could start with an open cover of $P^2(R)$ and pull it back to $R^3$. Then you see that this is equivalent to ask that $S^2$ is compact which follows from the Heine-Borel property for example. Mar3 comment Help showing homeomorphism If a map is bijective then being closed and being open are equivalent. This is just as you an define a map to be continuous using closed or open sets. Mar3 comment Help showing homeomorphism do you know that (if not try to prove it) bijective continuous maps $f: X \to Y$ are automatically homeomorphisms provided that $X$ is quasicompact and $Y$ is a Hausdorff-space? A hint is - try to prove that $f$ is closed instead of open. Feb23 comment Is the map $S^{2n+1}\rightarrow \mathbb{C}P^n \rightarrow \mathbb{C}P^n/\mathbb{C}P^{n-1}\cong S^{2n}$ essential? yes. you are right :) Feb22 comment Orientable double covers for non-orientable manifolds You are welcome. Were you able to check all details? Feb22 comment Is the map $S^{2n+1}\rightarrow \mathbb{C}P^n \rightarrow \mathbb{C}P^n/\mathbb{C}P^{n-1}\cong S^{2n}$ essential? I am sorry for being suspicious. I think the fact that the map is null-homotopic for $n$ even is easier. In any case you have this map $f: S^{2n+1} \to S^{2n}$ and if $n \geq 3$ this is a stable map, hence either zero or $\eta$. Hence the Cone of this map is either $\Sigma^n(\mathbb{C}P^2)$ or $S^{2n+2}\vee S^{2n}$. In both cases you can detect the homotopy type of that space by $Sq^2$ which is either zero or not. So I think for $n$ even the map you describe is null-homotopic for these simple reasons. Feb20 answered Orientable double covers for non-orientable manifolds Feb18 comment Some questions about $S^n$ Concerning the complex structures, there is a $K$-theoretic proof for the cases $n > 4$ and $n \neq 6$. Then in the case $n=4$ usually one does another argument. This case I think is actually much easier. Assume $S^4$ where almost complex. Then you can compute the first Pontryagin class in terms of chern classes and get a contradiction (the pontryagin class vanishes, but the second chern class is the euler class, so cannot be evaluated to zero). Feb18 answered Tangent Bundle of a Riemann Surface Feb11 revised cohomology groups and the pontryagin construction added 1 characters in body Feb11 answered cohomology groups and the pontryagin construction Feb4 comment Chern classes of free quotient manoflds at least rationally it does give you some information right? the problem is only that in general we do not know what $p^*(c_i(X/G))$ is since we don't know what the projection $p$ induces on cohomology. But rationally $p^*: H^*(X/G;\mathbb{Q}) \to H^*(X;\mathbb{Q})$ is just the inclusion of the $G$-invariants, right? Similarly, Chern Weyl theory should also tell you that (using an equivariant connection) you should be able to compute the chern classes quite explicitely? Jan27 awarded Fanatic Jan24 answered help-need to determine that this induced map is the zero map Jan18 comment What is the difference between homotopy and homeomorphism? Yes, but I did not want to mention it in this way in order to stay close to the question :)