homeomorphic image of an simply connected set
$begingroup$
My Question:
How can I prove, that the homeomorphic image of an simply connected set is also simply connected?
It is
konvex $ rightarrow $ radially $ rightarrow $ simply connected.
I have no idea how to prove that.
Appreciating any help! :-)
real-analysis general-topology
asked Jan 16 at 17:27
wondering1123wondering1123
14911
$endgroup$
add a comment |
$begingroup$
My Question:
How can I prove, that the homeomorphic image of an simply connected set is also simply connected?
It is
konvex $ rightarrow $ radially $ rightarrow $ simply connected.
I have no idea how to prove that.
Appreciating any help! :-)
real-analysis general-topology
asked Jan 16 at 17:27
wondering1123wondering1123
14911
$endgroup$
$begingroup$
An homeomorphism induces isomorphism in homotopy (fundamental group), whence the thesis.
$endgroup$
– Blumer
Jan 16 at 17:31
add a comment |
$begingroup$
My Question:
How can I prove, that the homeomorphic image of an simply connected set is also simply connected?
It is
konvex $ rightarrow $ radially $ rightarrow $ simply connected.
I have no idea how to prove that.
Appreciating any help! :-)
real-analysis general-topology
asked Jan 16 at 17:27
wondering1123wondering1123
14911
$endgroup$
My Question:
How can I prove, that the homeomorphic image of an simply connected set is also simply connected?
It is
konvex $ rightarrow $ radially $ rightarrow $ simply connected.
I have no idea how to prove that.
Appreciating any help! :-)
real-analysis general-topology
real-analysis general-topology
asked Jan 16 at 17:27
wondering1123wondering1123
14911
asked Jan 16 at 17:27
wondering1123wondering1123
14911
asked Jan 16 at 17:27
wondering1123wondering1123
14911
asked Jan 16 at 17:27
wondering1123wondering1123
14911
asked Jan 16 at 17:27
wondering1123wondering1123
14911
14911
$begingroup$
An homeomorphism induces isomorphism in homotopy (fundamental group), whence the thesis.
$endgroup$
– Blumer
Jan 16 at 17:31
add a comment |
$begingroup$
An homeomorphism induces isomorphism in homotopy (fundamental group), whence the thesis.
$endgroup$
– Blumer
Jan 16 at 17:31
$begingroup$
An homeomorphism induces isomorphism in homotopy (fundamental group), whence the thesis.
$endgroup$
– Blumer
Jan 16 at 17:31
$begingroup$
An homeomorphism induces isomorphism in homotopy (fundamental group), whence the thesis.
$endgroup$
– Blumer
Jan 16 at 17:31
add a comment |
1 Answer
1
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oldest
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$begingroup$
Homeomorphisms preserve all topological aspects of a space, essentially by definition. Concretely, assume $h:X to Y$ is a homeomorphism and $X$ is simply connected. This means that $X$ is path-connected and that all the loops in $X$ are contractible.
To see that $Y$ is path-connected, pick any two points $y_1,y_2 in Y$. Then since $X$ is path-connected, there exists a path $p:I to X$ (where $I = [0,1]$) starting at $h^{-1}(y_1)$ and ending at $h^{-1}(y_2)$. Then $h circ p:I to Y$ is a path starting at $y_1$ and ending at $y_2$.
To see that any loop in $Y$ is contractible, pick a loop $l:S^1 to Y$ based at a point $y_0 in Y$. Then $h^{-1}circ l:S^1to X$ is a loop in $X$ based at $x_0 = h^{-1}(y_0) in X$, which is contractible by hypothesis. This means that there exists a homotopy $H:S^1 times I to X$ such that $H(t, 0) = h^{-1}(l(t))$ and $H(t,1) = x_0$. Then $h circ H:S^1 times I to Y$ provides the necessary homotopy that contracts $l$ to $y_0$.
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
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add a comment |
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$begingroup$
Homeomorphisms preserve all topological aspects of a space, essentially by definition. Concretely, assume $h:X to Y$ is a homeomorphism and $X$ is simply connected. This means that $X$ is path-connected and that all the loops in $X$ are contractible.
To see that $Y$ is path-connected, pick any two points $y_1,y_2 in Y$. Then since $X$ is path-connected, there exists a path $p:I to X$ (where $I = [0,1]$) starting at $h^{-1}(y_1)$ and ending at $h^{-1}(y_2)$. Then $h circ p:I to Y$ is a path starting at $y_1$ and ending at $y_2$.
To see that any loop in $Y$ is contractible, pick a loop $l:S^1 to Y$ based at a point $y_0 in Y$. Then $h^{-1}circ l:S^1to X$ is a loop in $X$ based at $x_0 = h^{-1}(y_0) in X$, which is contractible by hypothesis. This means that there exists a homotopy $H:S^1 times I to X$ such that $H(t, 0) = h^{-1}(l(t))$ and $H(t,1) = x_0$. Then $h circ H:S^1 times I to Y$ provides the necessary homotopy that contracts $l$ to $y_0$.
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
$endgroup$
add a comment |
$begingroup$
Homeomorphisms preserve all topological aspects of a space, essentially by definition. Concretely, assume $h:X to Y$ is a homeomorphism and $X$ is simply connected. This means that $X$ is path-connected and that all the loops in $X$ are contractible.
To see that $Y$ is path-connected, pick any two points $y_1,y_2 in Y$. Then since $X$ is path-connected, there exists a path $p:I to X$ (where $I = [0,1]$) starting at $h^{-1}(y_1)$ and ending at $h^{-1}(y_2)$. Then $h circ p:I to Y$ is a path starting at $y_1$ and ending at $y_2$.
To see that any loop in $Y$ is contractible, pick a loop $l:S^1 to Y$ based at a point $y_0 in Y$. Then $h^{-1}circ l:S^1to X$ is a loop in $X$ based at $x_0 = h^{-1}(y_0) in X$, which is contractible by hypothesis. This means that there exists a homotopy $H:S^1 times I to X$ such that $H(t, 0) = h^{-1}(l(t))$ and $H(t,1) = x_0$. Then $h circ H:S^1 times I to Y$ provides the necessary homotopy that contracts $l$ to $y_0$.
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
$endgroup$
add a comment |
$begingroup$
Homeomorphisms preserve all topological aspects of a space, essentially by definition. Concretely, assume $h:X to Y$ is a homeomorphism and $X$ is simply connected. This means that $X$ is path-connected and that all the loops in $X$ are contractible.
To see that $Y$ is path-connected, pick any two points $y_1,y_2 in Y$. Then since $X$ is path-connected, there exists a path $p:I to X$ (where $I = [0,1]$) starting at $h^{-1}(y_1)$ and ending at $h^{-1}(y_2)$. Then $h circ p:I to Y$ is a path starting at $y_1$ and ending at $y_2$.
To see that any loop in $Y$ is contractible, pick a loop $l:S^1 to Y$ based at a point $y_0 in Y$. Then $h^{-1}circ l:S^1to X$ is a loop in $X$ based at $x_0 = h^{-1}(y_0) in X$, which is contractible by hypothesis. This means that there exists a homotopy $H:S^1 times I to X$ such that $H(t, 0) = h^{-1}(l(t))$ and $H(t,1) = x_0$. Then $h circ H:S^1 times I to Y$ provides the necessary homotopy that contracts $l$ to $y_0$.
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
$endgroup$
Homeomorphisms preserve all topological aspects of a space, essentially by definition. Concretely, assume $h:X to Y$ is a homeomorphism and $X$ is simply connected. This means that $X$ is path-connected and that all the loops in $X$ are contractible.
To see that $Y$ is path-connected, pick any two points $y_1,y_2 in Y$. Then since $X$ is path-connected, there exists a path $p:I to X$ (where $I = [0,1]$) starting at $h^{-1}(y_1)$ and ending at $h^{-1}(y_2)$. Then $h circ p:I to Y$ is a path starting at $y_1$ and ending at $y_2$.
To see that any loop in $Y$ is contractible, pick a loop $l:S^1 to Y$ based at a point $y_0 in Y$. Then $h^{-1}circ l:S^1to X$ is a loop in $X$ based at $x_0 = h^{-1}(y_0) in X$, which is contractible by hypothesis. This means that there exists a homotopy $H:S^1 times I to X$ such that $H(t, 0) = h^{-1}(l(t))$ and $H(t,1) = x_0$. Then $h circ H:S^1 times I to Y$ provides the necessary homotopy that contracts $l$ to $y_0$.
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
answered Jan 16 at 21:08
Alex ProvostAlex Provost
15.6k32351
15.6k32351
add a comment |
add a comment |
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$begingroup$
An homeomorphism induces isomorphism in homotopy (fundamental group), whence the thesis.
$endgroup$
– Blumer
Jan 16 at 17:31