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5h
answered Prove a square is homeomorphic to a circle
5h
comment Any homeomorphism from $D^2$ to $D^2$ maps $\partial D^2$ onto $\partial D^2$
What do you mean? The identity obviously satisfies what I'm trying to prove.
6h
asked Any homeomorphism from $D^2$ to $D^2$ maps $\partial D^2$ onto $\partial D^2$
Aug
29
comment CW complex structure of the projective space $\mathbb{RP}^n$
By the "natural inclusion of the hemisphere $D^n\to S^n$" you mean the map $(x_1,...,x_n)\mapsto(x_1,...,x_n,\sqrt{1-x_1^2-x_2^2-...-x_n^2})$ or the map $tu\mapsto (\sin(t\pi/2)u,\cos(t\pi/2))$ where $t\in [0,1]$ and $u\in S^{n-1}$?.
Aug
28
comment What is the Quotient on the Coproduct in Adjunction Spaces
How do you get $f(a)\tilde{} a$ from there? Also the third line of the equivalence relation looks very weird.
Aug
25
revised oThe cofinite topology is the infimum of the hausdorff topologies in a set $X$
edited body; edited title
Aug
23
comment Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
Ok I think I figured it out. It's true that $p$ is not an open map as your example shows, but the reason $p$ is closed is not because multiplication $\cdot:R\times R\to R$ is a closed map (consider the set $\{(n,1/n^2):n\in N\}$ whose image is $\{1/n:n\in N\}$ that is not closed in $R$) but because $S^1\times I$ is compact and $D^2$ is hausdorff. What do you think?
Aug
23
revised Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
edited title
Aug
23
comment Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
It's not that I can't prove that $p$ is a surjection, that's trivial, I can't prove $p$ is an identification (I actually suspect it's an open map (this implies identification), but I'm not sure). If you know why it's an open map, please elaborate on that.
Aug
23
comment Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
Check Dugundji's book pages 123 and 124: Transgression theorem about identifications, this is the very same situation.
Aug
23
comment Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
$Fp^{-1}(p(x))=F(x)$, i.e. one has to prove that $p(x)=p(x')\implies F(x)=F(x')$, so that $Fp^{-1}$ is indeed well defined.
Aug
23
comment Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
$p^{-1}$ is the inverse image of $U$, i.e. $p^{-1}(U)=\{x\in X:p(x)\in U\}$.
Aug
23
comment Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
By identification I mean a surjection $p:X\to Y$ such that $U$ is open in $Y$ iff $p^{-1}(U)$ is open in $X$, not a bijection.
Aug
23
revised Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
added 2 characters in body
Aug
23
asked Prove $S^1\times I\to D^2,(s,t)\mapsto ts$ is an identification.
Aug
22
accepted oThe cofinite topology is the infimum of the hausdorff topologies in a set $X$
Aug
22
revised oThe cofinite topology is the infimum of the hausdorff topologies in a set $X$
added 19 characters in body
Aug
22
asked oThe cofinite topology is the infimum of the hausdorff topologies in a set $X$
Aug
14
accepted Choosing a continuous function satisfying the mean value theorem
Aug
13
asked Choosing a continuous function satisfying the mean value theorem