Mixing
Properties of covering spaces replacing base points by contractible subspaces
$begingroup$
Let $(X,A)$ and $(Y,B)$ pairs satisfying the homotopy extension property (or a CW-pairs if necessary) with $A$ and $B$ contractible. Let $f:(X,A)to (Y,B)$ a map of pairs and let $p_X:widetilde{X}to X$ and $p_Y:widetilde{Y}to Y$ covering maps.
By definition of covering space, every point $xin X$ has a neighbourhood $U$ such that $p^{-1}(U)$ is a disjoint union of open sets each of them mapped homeomorphically onto $U$ by $p$, so in particular $p^{-1}(x)$ is a discrete set with one point on each component of $p^{-1}(U)$. By the theory of covering spaces we know that under certain conditions, a map $(X,x_0)to (Y,y_0)$ may lift to $(widetilde{X},tilde{x}_0)to (widetilde{Y},tilde{y}_0)$.
I am interested in knowing if such property are satisfied in my setting, namely, is $p^{-1}(A)$ a disjoint union of subspaces of $widetilde{X}$ each of them mapped homeomorphically by $p$ onto $A$.
If so, I can choose $widetilde{A}$ and $widetilde{B}$ to be such subspaces of $widetilde{X}$ and $widetilde{Y}$ respectively. Then I wonder if the map $f:(X,A)to (Y,B)$ can be (uniquely) lifted to $tilde{f}:(widetilde{X},widetilde{A})to (widetilde{Y}, widetilde{B})$ making the diagram of maps of pairs and covering maps commute.
algebraic-topology homotopy-theory covering-spaces cw-complexes homotopy-extension-property
asked Jan 27 at 19:38
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
Let $(X,A)$ and $(Y,B)$ pairs satisfying the homotopy extension property (or a CW-pairs if necessary) with $A$ and $B$ contractible. Let $f:(X,A)to (Y,B)$ a map of pairs and let $p_X:widetilde{X}to X$ and $p_Y:widetilde{Y}to Y$ covering maps.
By definition of covering space, every point $xin X$ has a neighbourhood $U$ such that $p^{-1}(U)$ is a disjoint union of open sets each of them mapped homeomorphically onto $U$ by $p$, so in particular $p^{-1}(x)$ is a discrete set with one point on each component of $p^{-1}(U)$. By the theory of covering spaces we know that under certain conditions, a map $(X,x_0)to (Y,y_0)$ may lift to $(widetilde{X},tilde{x}_0)to (widetilde{Y},tilde{y}_0)$.
I am interested in knowing if such property are satisfied in my setting, namely, is $p^{-1}(A)$ a disjoint union of subspaces of $widetilde{X}$ each of them mapped homeomorphically by $p$ onto $A$.
If so, I can choose $widetilde{A}$ and $widetilde{B}$ to be such subspaces of $widetilde{X}$ and $widetilde{Y}$ respectively. Then I wonder if the map $f:(X,A)to (Y,B)$ can be (uniquely) lifted to $tilde{f}:(widetilde{X},widetilde{A})to (widetilde{Y}, widetilde{B})$ making the diagram of maps of pairs and covering maps commute.
algebraic-topology homotopy-theory covering-spaces cw-complexes homotopy-extension-property
asked Jan 27 at 19:38
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
Let $(X,A)$ and $(Y,B)$ pairs satisfying the homotopy extension property (or a CW-pairs if necessary) with $A$ and $B$ contractible. Let $f:(X,A)to (Y,B)$ a map of pairs and let $p_X:widetilde{X}to X$ and $p_Y:widetilde{Y}to Y$ covering maps.
By definition of covering space, every point $xin X$ has a neighbourhood $U$ such that $p^{-1}(U)$ is a disjoint union of open sets each of them mapped homeomorphically onto $U$ by $p$, so in particular $p^{-1}(x)$ is a discrete set with one point on each component of $p^{-1}(U)$. By the theory of covering spaces we know that under certain conditions, a map $(X,x_0)to (Y,y_0)$ may lift to $(widetilde{X},tilde{x}_0)to (widetilde{Y},tilde{y}_0)$.
I am interested in knowing if such property are satisfied in my setting, namely, is $p^{-1}(A)$ a disjoint union of subspaces of $widetilde{X}$ each of them mapped homeomorphically by $p$ onto $A$.
If so, I can choose $widetilde{A}$ and $widetilde{B}$ to be such subspaces of $widetilde{X}$ and $widetilde{Y}$ respectively. Then I wonder if the map $f:(X,A)to (Y,B)$ can be (uniquely) lifted to $tilde{f}:(widetilde{X},widetilde{A})to (widetilde{Y}, widetilde{B})$ making the diagram of maps of pairs and covering maps commute.
algebraic-topology homotopy-theory covering-spaces cw-complexes homotopy-extension-property
asked Jan 27 at 19:38
JaviJavi
3,0212832
$endgroup$
Let $(X,A)$ and $(Y,B)$ pairs satisfying the homotopy extension property (or a CW-pairs if necessary) with $A$ and $B$ contractible. Let $f:(X,A)to (Y,B)$ a map of pairs and let $p_X:widetilde{X}to X$ and $p_Y:widetilde{Y}to Y$ covering maps.
By definition of covering space, every point $xin X$ has a neighbourhood $U$ such that $p^{-1}(U)$ is a disjoint union of open sets each of them mapped homeomorphically onto $U$ by $p$, so in particular $p^{-1}(x)$ is a discrete set with one point on each component of $p^{-1}(U)$. By the theory of covering spaces we know that under certain conditions, a map $(X,x_0)to (Y,y_0)$ may lift to $(widetilde{X},tilde{x}_0)to (widetilde{Y},tilde{y}_0)$.
I am interested in knowing if such property are satisfied in my setting, namely, is $p^{-1}(A)$ a disjoint union of subspaces of $widetilde{X}$ each of them mapped homeomorphically by $p$ onto $A$.
If so, I can choose $widetilde{A}$ and $widetilde{B}$ to be such subspaces of $widetilde{X}$ and $widetilde{Y}$ respectively. Then I wonder if the map $f:(X,A)to (Y,B)$ can be (uniquely) lifted to $tilde{f}:(widetilde{X},widetilde{A})to (widetilde{Y}, widetilde{B})$ making the diagram of maps of pairs and covering maps commute.
algebraic-topology homotopy-theory covering-spaces cw-complexes homotopy-extension-property
algebraic-topology homotopy-theory covering-spaces cw-complexes homotopy-extension-property
asked Jan 27 at 19:38
JaviJavi
3,0212832
asked Jan 27 at 19:38
JaviJavi
3,0212832
edited Jan 27 at 19:44
Javi
asked Jan 27 at 19:38
JaviJavi
3,0212832
asked Jan 27 at 19:38
JaviJavi
3,0212832
asked Jan 27 at 19:38
JaviJavi
3,0212832
3,0212832
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
In general a fibre bundle over a contractible space is trivial, so $p_X^{-1}(A)cong A times F_{x_0}$ where $F_{x_0}$ is the (in our case discrete) fibre over any point in $A$.
Wether or not a map $fcolon (X, A) to (Y, B)$ lifts to $tilde{f}colon(tilde{X}, tilde{A}) to (tilde{Y}, tilde{B})$ then doesn't really depend on $A$ or $B$ because they are homotopically trivial, so it's essentially the same lifting problem as if they were points.
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
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$begingroup$
Isn't it automatic ? (edit : I've actually assumed that in the decomposition $p_Y^{-1}(B) = displaystylecoprod_{iin I}B$, each of the $B$ was a connected component; and that the coverings were normal)
Indeed, consider the basepoints $x_0,y_0$ to be in $tilde{A},tilde{B}$ respectively; and consider a lift $g:(tilde{X},tilde{x_0})to (tilde{Y},tilde{y_0})$.
Then $g(tilde{A})$ is a connected subset of $tilde{Y}$ that contains $tilde{y_0}$. Moreover, $p_Y(g(tilde{A}))=f(p_X(tilde{A})) = f(A)subset B$. So $g(tilde{A})$ is included in $p_Y^{-1}(B)$, more specifically in the connected component containing $tilde{y_0}$, which is $tilde{B}$; so $f$ is actually a map of pairs.
Now if, say the baspoint $tilde{y_0}$ is not in $tilde{B}$, but in some other connected component of $p_Y^{-1}(B)$, then up to a deck transformation, you have the same result. So as long as the basepoints $x_0,y_0$ are in $A,B$ respectively, you can find such a lift. Now if $X,Y$ are path-connected, you can assume that this is the case.
So it suffices that the map can be lifted to a $g$ for it to be lifted to a map of pairs
answered Jan 27 at 20:23
MaxMax
15.6k11143
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
In general a fibre bundle over a contractible space is trivial, so $p_X^{-1}(A)cong A times F_{x_0}$ where $F_{x_0}$ is the (in our case discrete) fibre over any point in $A$.
Wether or not a map $fcolon (X, A) to (Y, B)$ lifts to $tilde{f}colon(tilde{X}, tilde{A}) to (tilde{Y}, tilde{B})$ then doesn't really depend on $A$ or $B$ because they are homotopically trivial, so it's essentially the same lifting problem as if they were points.
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
$endgroup$
add a comment |
$begingroup$
In general a fibre bundle over a contractible space is trivial, so $p_X^{-1}(A)cong A times F_{x_0}$ where $F_{x_0}$ is the (in our case discrete) fibre over any point in $A$.
Wether or not a map $fcolon (X, A) to (Y, B)$ lifts to $tilde{f}colon(tilde{X}, tilde{A}) to (tilde{Y}, tilde{B})$ then doesn't really depend on $A$ or $B$ because they are homotopically trivial, so it's essentially the same lifting problem as if they were points.
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
$endgroup$
add a comment |
$begingroup$
In general a fibre bundle over a contractible space is trivial, so $p_X^{-1}(A)cong A times F_{x_0}$ where $F_{x_0}$ is the (in our case discrete) fibre over any point in $A$.
Wether or not a map $fcolon (X, A) to (Y, B)$ lifts to $tilde{f}colon(tilde{X}, tilde{A}) to (tilde{Y}, tilde{B})$ then doesn't really depend on $A$ or $B$ because they are homotopically trivial, so it's essentially the same lifting problem as if they were points.
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
$endgroup$
In general a fibre bundle over a contractible space is trivial, so $p_X^{-1}(A)cong A times F_{x_0}$ where $F_{x_0}$ is the (in our case discrete) fibre over any point in $A$.
Wether or not a map $fcolon (X, A) to (Y, B)$ lifts to $tilde{f}colon(tilde{X}, tilde{A}) to (tilde{Y}, tilde{B})$ then doesn't really depend on $A$ or $B$ because they are homotopically trivial, so it's essentially the same lifting problem as if they were points.
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
answered Jan 27 at 19:50
WilliamWilliam
2,8201224
2,8201224
add a comment |
add a comment |
$begingroup$
Isn't it automatic ? (edit : I've actually assumed that in the decomposition $p_Y^{-1}(B) = displaystylecoprod_{iin I}B$, each of the $B$ was a connected component; and that the coverings were normal)
Indeed, consider the basepoints $x_0,y_0$ to be in $tilde{A},tilde{B}$ respectively; and consider a lift $g:(tilde{X},tilde{x_0})to (tilde{Y},tilde{y_0})$.
Then $g(tilde{A})$ is a connected subset of $tilde{Y}$ that contains $tilde{y_0}$. Moreover, $p_Y(g(tilde{A}))=f(p_X(tilde{A})) = f(A)subset B$. So $g(tilde{A})$ is included in $p_Y^{-1}(B)$, more specifically in the connected component containing $tilde{y_0}$, which is $tilde{B}$; so $f$ is actually a map of pairs.
Now if, say the baspoint $tilde{y_0}$ is not in $tilde{B}$, but in some other connected component of $p_Y^{-1}(B)$, then up to a deck transformation, you have the same result. So as long as the basepoints $x_0,y_0$ are in $A,B$ respectively, you can find such a lift. Now if $X,Y$ are path-connected, you can assume that this is the case.
So it suffices that the map can be lifted to a $g$ for it to be lifted to a map of pairs
answered Jan 27 at 20:23
MaxMax
15.6k11143
$endgroup$
add a comment |
$begingroup$
Isn't it automatic ? (edit : I've actually assumed that in the decomposition $p_Y^{-1}(B) = displaystylecoprod_{iin I}B$, each of the $B$ was a connected component; and that the coverings were normal)
Indeed, consider the basepoints $x_0,y_0$ to be in $tilde{A},tilde{B}$ respectively; and consider a lift $g:(tilde{X},tilde{x_0})to (tilde{Y},tilde{y_0})$.
Then $g(tilde{A})$ is a connected subset of $tilde{Y}$ that contains $tilde{y_0}$. Moreover, $p_Y(g(tilde{A}))=f(p_X(tilde{A})) = f(A)subset B$. So $g(tilde{A})$ is included in $p_Y^{-1}(B)$, more specifically in the connected component containing $tilde{y_0}$, which is $tilde{B}$; so $f$ is actually a map of pairs.
Now if, say the baspoint $tilde{y_0}$ is not in $tilde{B}$, but in some other connected component of $p_Y^{-1}(B)$, then up to a deck transformation, you have the same result. So as long as the basepoints $x_0,y_0$ are in $A,B$ respectively, you can find such a lift. Now if $X,Y$ are path-connected, you can assume that this is the case.
So it suffices that the map can be lifted to a $g$ for it to be lifted to a map of pairs
answered Jan 27 at 20:23
MaxMax
15.6k11143
$endgroup$
add a comment |
$begingroup$
Isn't it automatic ? (edit : I've actually assumed that in the decomposition $p_Y^{-1}(B) = displaystylecoprod_{iin I}B$, each of the $B$ was a connected component; and that the coverings were normal)
Indeed, consider the basepoints $x_0,y_0$ to be in $tilde{A},tilde{B}$ respectively; and consider a lift $g:(tilde{X},tilde{x_0})to (tilde{Y},tilde{y_0})$.
Then $g(tilde{A})$ is a connected subset of $tilde{Y}$ that contains $tilde{y_0}$. Moreover, $p_Y(g(tilde{A}))=f(p_X(tilde{A})) = f(A)subset B$. So $g(tilde{A})$ is included in $p_Y^{-1}(B)$, more specifically in the connected component containing $tilde{y_0}$, which is $tilde{B}$; so $f$ is actually a map of pairs.
Now if, say the baspoint $tilde{y_0}$ is not in $tilde{B}$, but in some other connected component of $p_Y^{-1}(B)$, then up to a deck transformation, you have the same result. So as long as the basepoints $x_0,y_0$ are in $A,B$ respectively, you can find such a lift. Now if $X,Y$ are path-connected, you can assume that this is the case.
So it suffices that the map can be lifted to a $g$ for it to be lifted to a map of pairs
answered Jan 27 at 20:23
MaxMax
15.6k11143
$endgroup$
Isn't it automatic ? (edit : I've actually assumed that in the decomposition $p_Y^{-1}(B) = displaystylecoprod_{iin I}B$, each of the $B$ was a connected component; and that the coverings were normal)
Indeed, consider the basepoints $x_0,y_0$ to be in $tilde{A},tilde{B}$ respectively; and consider a lift $g:(tilde{X},tilde{x_0})to (tilde{Y},tilde{y_0})$.
Then $g(tilde{A})$ is a connected subset of $tilde{Y}$ that contains $tilde{y_0}$. Moreover, $p_Y(g(tilde{A}))=f(p_X(tilde{A})) = f(A)subset B$. So $g(tilde{A})$ is included in $p_Y^{-1}(B)$, more specifically in the connected component containing $tilde{y_0}$, which is $tilde{B}$; so $f$ is actually a map of pairs.
Now if, say the baspoint $tilde{y_0}$ is not in $tilde{B}$, but in some other connected component of $p_Y^{-1}(B)$, then up to a deck transformation, you have the same result. So as long as the basepoints $x_0,y_0$ are in $A,B$ respectively, you can find such a lift. Now if $X,Y$ are path-connected, you can assume that this is the case.
So it suffices that the map can be lifted to a $g$ for it to be lifted to a map of pairs
answered Jan 27 at 20:23
MaxMax
15.6k11143
answered Jan 27 at 20:23
MaxMax
15.6k11143
answered Jan 27 at 20:23
MaxMax
15.6k11143
answered Jan 27 at 20:23
MaxMax
15.6k11143
15.6k11143
add a comment |
add a comment |
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