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edit approved Oct 23, 2017 at 18:54
Kyle Miller
edit approved Oct 23, 2017 at 18:54
Kyle Miller
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Homotopy equvalenceequivalence from contractibility of fiber.

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Suppose $X$ and $Y$ are two $CW$ complexes and $f:X\rightarrow Y$ is a continuous surjection such that fiber of each point (i.e. $f^{-1}(y)$ for each $y\in Y$) is contractible. Does it implies that $X$ and $Y$ are homotopy equivalent.

PS-1:By Whitehead's Theorem it will be enough to show that $f$ induces an isomorphism between all homotopy groups.

PS-2:In question Equivariant Cohomology for actions with finite stabilizersEquivariant Cohomology for actions with finite stabilizers there are some discussion regarding the above question but in terms of homology. If anybody thinks that my question can be a consequence of this discussion please explain the connection.

Suppose $X$ and $Y$ are two $CW$ complexes and $f:X\rightarrow Y$ is a continuous surjection such that fiber of each point (i.e. $f^{-1}(y)$ for each $y\in Y$) is contractible. Does it implies that $X$ and $Y$ are homotopy equivalent.

PS-1:By Whitehead's Theorem it will be enough to show that $f$ induces an isomorphism between all homotopy groups.

PS-2:In question Equivariant Cohomology for actions with finite stabilizers there are some discussion regarding the above question but in terms of homology. If anybody thinks that my question can be a consequence of this discussion please explain the connection.

Suppose $X$ and $Y$ are two $CW$ complexes and $f:X\rightarrow Y$ is a continuous surjection such that fiber of each point (i.e. $f^{-1}(y)$ for each $y\in Y$) is contractible. Does it implies that $X$ and $Y$ are homotopy equivalent.

PS-1:By Whitehead's Theorem it will be enough to show that $f$ induces an isomorphism between all homotopy groups.

PS-2:In question Equivariant Cohomology for actions with finite stabilizers there are some discussion regarding the above question but in terms of homology. If anybody thinks that my question can be a consequence of this discussion please explain the connection.

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Homotopy equvalence from contractibility of fiber.

Suppose $X$ and $Y$ are two $CW$ complexes and $f:X\rightarrow Y$ is a continuous surjection such that fiber of each point (i.e. $f^{-1}(y)$ for each $y\in Y$) is contractible. Does it implies that $X$ and $Y$ are homotopy equivalent.

PS-1:By Whitehead's Theorem it will be enough to show that $f$ induces an isomorphism between all homotopy groups.

PS-2:In question Equivariant Cohomology for actions with finite stabilizers there are some discussion regarding the above question but in terms of homology. If anybody thinks that my question can be a consequence of this discussion please explain the connection.