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Apr
22
comment The automorphism group of the real line with standard topology
@IttayWeiss I'm not talking about deformation retractions that respect the group structure, only about deformation retractions of the underlying topological space, I don't think the dynamic properties of the automorphisms plays any role in that. As for something that is compatible with the group structure, how about scaling in the domain? What about the maps $H:A\times[0,1]\to A, (f,t)\mapsto \lbrace x\mapsto\frac{f(tx)-f(0)}{t}+f(0)\rbrace$? At least when one considers the subgroup of $C^1$ diffeomorphisms, this should give a strong deformation retraction onto the affine functions.
Apr
22
comment The automorphism group of the real line with standard topology
@IttayWeiss I was only talking about the topology of this space, not the group structure. I remember proving that it is a topological group with the compact open topology years ago. don't you think that the space should strongly deformation retract onto the subset of all affine homeomorphisms with slope $\pm1$? From there it will deformation retract onto $\lbrace \pm1\rbrace$ I think...
Apr
22
comment The automorphism group of the real line with standard topology
It seems to me, at first sight, reasonable, that there should be two path components, and a strong deformation retraction onto $\lbrace 1,-1\rbrace$, where $1$ is shorthand for the identity of $\Bbb R$. After all, the homeomorphisms of $\Bbb R$ are precisely the surjective stricly monotone functions, and there are obvious "convex homotopies".
Apr
21
answered the mapping class group of the disk is trivial proof
Apr
21
revised Good topologies on $\mathcal{P}(X)$
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Apr
21
asked Good topologies on $\mathcal{P}(X)$
Apr
20
comment Bounded linear functionals on $L^\infty$.
Are you sure $C^0[0,1]$ isn't the space of continuous functions on the unit interval? The notation is standard, and they are all measurable and bounded.
Apr
20
comment Can a cube always be fitted into the projection of a cube?
Are you allowing for all projections, or only for the orthogonal ones (i.e. those projections with orthogonal image and kernel)?
Apr
20
comment Homology Whitehead theorem for non simply connected spaces
Ok, if you say so ^^ I awarded you the bounty, I guess I need to read the articles more in depth to believe their claims and my own explanation ^^
Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
comment Homology Whitehead theorem for non simply connected spaces
Also, are the spaces $X$ and $Z$ the actual spaces he uses to find homology equivalent (in $\pi_1$ and $H_*$), non homotopy equivalent spaces?
Apr
20
comment Homology Whitehead theorem for non simply connected spaces
@studiosus But as I understand it, it is only required that the Euler characteristics be equal and $m=1$, and it seems to me he is saying that $m=1$ whenever $\chi>\chi_{\min}(G,2)$. Since we know by example (the complex associated to the presentation $\langle a,b\mid a^3=b^2\rangle$) that there is a $[G,2]$-complex with $\chi=0$ we have $0\geq\chi_{\min}(G,2)$, and so, for any $[G,2]$-complex with $\chi\geq 1$, $m=1$. My point is that it apparantly doesn't matter that $\chi_{\min}(G,2)=0$ or not, it only matters that it be less than our $\chi$. Am I correctly interpreting the article?
Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
comment Homology Whitehead theorem for non simply connected spaces
I left an answer to tell you what I understood from your examples, could you take a look at it and correct my understanding of it?
Apr
20
answered Homology Whitehead theorem for non simply connected spaces
Apr
20
answered Is $\sin(\arcsin(x))$ equal to $x$?
Apr
20
comment Euler characteristic and free action
What action of $G$ on $K^2$ are you considering?