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13
awarded  Notable Question
Jul
30
awarded  Popular Question
Jul
2
accepted Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
Jul
2
comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
Thank you, nice work. Since the most work was done, I'll accept your answer. Only detail I wanted to tweak is that in my second comment here equality $f(e_1\cdot[w])=f(e_2)$ doesn't hold, but this doesn't affect surjectivity I was about to prove anyway.
Jul
2
comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
In order to prove surjectivity, let $e\in E$. Pick any other $e_1 \in E$ and let $f(e_1)=e_2$. Now let $w$ be the path between $e_2$ and $e$ (exists, since $E$ path-connected). Then $e = e_2\cdot[w] = f(e_1)\cdot[w] = f(e_1 \cdot [w]) = f(e_2)$. So $f$ is surjective, and bijective as it maps into itself. Does this look sound?
Jul
2
comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
Concerning bijectivity: in my another question where you also engaged in discussion I have defined an action of the fundamental group on $E$ (path (homotopy class) in $X$ maps each point in $E$ to the endpoint of the unique lifted path) and proved that $f(e\cdot [w]) = f(e) \cdot[w]$, where $f$ is a continuous map on $E$ having $pf=p$ and $e\cdot[w]$ is aforementioned action.
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
1
comment Deck transformation acting properly discontinuously assumed covering space is path-connected
Thank you for an answer. But I still don't understand one thing (which possibly prevented me from solving the problem myself): why can't we take another neighborhood, in your example, one that would contain both $z \neq u$ from different copies? Is it because they are disjoint?
Jul
1
revised Deck transformation acting properly discontinuously assumed covering space is path-connected
edited tags
Jul
1
asked Deck transformation acting properly discontinuously assumed covering space is path-connected
Jul
1
comment Covering through group action and corresponding deck transformations
@squirrel Thanks a lot for a detailed explanation! So without $(*)$ statement $\bar{\phi}$ wouldn't be well-defined, right?
Jul
1
comment Covering through group action and corresponding deck transformations
@squirrel Sorry, could you explain the implication in line 4, where you from $(*)$ imply deck transformation? You use it seemingly in $G(gH_e)=G(H_e)$, but I don't see any connection behind this.
Jul
1
comment Covering through group action and corresponding deck transformations
Concerning (i): by this logic one could argue in the opposite direction, because equally $G \backslash E$ is a quotient and $E\rightarrow G\backslash E$ is a covering, thus getting $G \backslash E$ is a covering of $H \backslash E$, or am I missing something?
Jun
30
comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
This is how it's written on my paper. Ok, I reread this and this really sounds a bit misleading. Following was meant: Let $p$ and $f$ be as given above and $pf = p$. Then $f$ is a homeo.
Jun
30
comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
I guess spacebook agrees with me on that.
Jun
30
comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
I thought topologist's comb is path connected but not locally path connected.
Jun
30
asked Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
May
11
accepted How to fit parameters to given data of given function?
May
11
asked How to fit parameters to given data of given function?