homotopy and (co)filtered limits
Suppose we have a (co)filtered digaram $dots rightarrow X_{2}rightarrow X_{1}$ of topological space. Is is true that the natural map $pi_{0}[lim X_{i}]rightarrow lim pi_{0}(X_{i})$ is an isomorphism ?
at.algebraic-topology homotopy-theory
edited Dec 7 at 14:58
Rad80
1032
asked Dec 6 at 20:23
Ofra
582520
add a comment |
Suppose we have a (co)filtered digaram $dots rightarrow X_{2}rightarrow X_{1}$ of topological space. Is is true that the natural map $pi_{0}[lim X_{i}]rightarrow lim pi_{0}(X_{i})$ is an isomorphism ?
at.algebraic-topology homotopy-theory
edited Dec 7 at 14:58
Rad80
1032
asked Dec 6 at 20:23
Ofra
582520
7
No. For example, take all your $X_i$ to be the $1$-sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected.
– Achim Krause
Dec 6 at 21:20
add a comment |
Suppose we have a (co)filtered digaram $dots rightarrow X_{2}rightarrow X_{1}$ of topological space. Is is true that the natural map $pi_{0}[lim X_{i}]rightarrow lim pi_{0}(X_{i})$ is an isomorphism ?
at.algebraic-topology homotopy-theory
edited Dec 7 at 14:58
Rad80
1032
asked Dec 6 at 20:23
Ofra
582520
Suppose we have a (co)filtered digaram $dots rightarrow X_{2}rightarrow X_{1}$ of topological space. Is is true that the natural map $pi_{0}[lim X_{i}]rightarrow lim pi_{0}(X_{i})$ is an isomorphism ?
at.algebraic-topology homotopy-theory
at.algebraic-topology homotopy-theory
edited Dec 7 at 14:58
Rad80
1032
asked Dec 6 at 20:23
Ofra
582520
edited Dec 7 at 14:58
Rad80
1032
asked Dec 6 at 20:23
Ofra
582520
edited Dec 7 at 14:58
Rad80
1032
edited Dec 7 at 14:58
Rad80
1032
edited Dec 7 at 14:58
Rad80
1032
1032
asked Dec 6 at 20:23
Ofra
582520
asked Dec 6 at 20:23
Ofra
582520
asked Dec 6 at 20:23
Ofra
582520
582520
7
No. For example, take all your $X_i$ to be the $1$-sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected.
– Achim Krause
Dec 6 at 21:20
add a comment |
7
No. For example, take all your $X_i$ to be the $1$-sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected.
– Achim Krause
Dec 6 at 21:20
7
7
No. For example, take all your $X_i$ to be the $1$-sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected.
– Achim Krause
Dec 6 at 21:20
No. For example, take all your $X_i$ to be the $1$-sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected.
– Achim Krause
Dec 6 at 21:20
add a comment |
1 Answer
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active
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This is not true, for two distinct reasons.
The first is that the inverse system of spaces may not behave well homotopy-theoretically. If $X_n = [n, infty) subset Bbb R$, then the limit of $dots to X_2 to X_1 to X_0$ is empty. However, on path components it is the constant system $dots to * to * to *$, with limit $*$. Roughly, you might have a path component that is represented by any space $X_i$ that is not represented by any compatible system of points. This might make $pi_0 lim X_i to lim pi_0 X_i$ not surjective.
The second is the opposite: the map $pi_0 lim X_i to lim pi_0 X_i$ may not be injective. In this case, you may have two points $x$ and $y$ in the limit such that the images in any individual $x_i$ are connected by a path, but where no path can be compatibly lifted all the way up the tower. For example, if $f:S^1 to S^1$ is a degree-2 covering map, then the limit of the tower $$dots xrightarrow{f} S^1 xrightarrow{f} S^1 xrightarrow{f} S^1$$ is called the 2-adic solenoid and it has uncountably many path components.
The first problem goes away if the maps $X_i to X_{i-1}$ are fibrations, and in this case we often call the limit a homotopy limit. The second problem does not go away in this case, but Milnor proved that $pi_0 lim X_i$ is built out of two terms: $lim(pi_0 X_i)$ and a second term called $lim^1(pi_1 X_i)$. In particular, if the spaces $X_i$ are simply-connected then there are no contributions from the second term, and so there will be an isomorphism $pi_0(lim X_i) to lim pi_0(X_i)$.
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
add a comment |
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This is not true, for two distinct reasons.
The first is that the inverse system of spaces may not behave well homotopy-theoretically. If $X_n = [n, infty) subset Bbb R$, then the limit of $dots to X_2 to X_1 to X_0$ is empty. However, on path components it is the constant system $dots to * to * to *$, with limit $*$. Roughly, you might have a path component that is represented by any space $X_i$ that is not represented by any compatible system of points. This might make $pi_0 lim X_i to lim pi_0 X_i$ not surjective.
The second is the opposite: the map $pi_0 lim X_i to lim pi_0 X_i$ may not be injective. In this case, you may have two points $x$ and $y$ in the limit such that the images in any individual $x_i$ are connected by a path, but where no path can be compatibly lifted all the way up the tower. For example, if $f:S^1 to S^1$ is a degree-2 covering map, then the limit of the tower $$dots xrightarrow{f} S^1 xrightarrow{f} S^1 xrightarrow{f} S^1$$ is called the 2-adic solenoid and it has uncountably many path components.
The first problem goes away if the maps $X_i to X_{i-1}$ are fibrations, and in this case we often call the limit a homotopy limit. The second problem does not go away in this case, but Milnor proved that $pi_0 lim X_i$ is built out of two terms: $lim(pi_0 X_i)$ and a second term called $lim^1(pi_1 X_i)$. In particular, if the spaces $X_i$ are simply-connected then there are no contributions from the second term, and so there will be an isomorphism $pi_0(lim X_i) to lim pi_0(X_i)$.
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
add a comment |
This is not true, for two distinct reasons.
The first is that the inverse system of spaces may not behave well homotopy-theoretically. If $X_n = [n, infty) subset Bbb R$, then the limit of $dots to X_2 to X_1 to X_0$ is empty. However, on path components it is the constant system $dots to * to * to *$, with limit $*$. Roughly, you might have a path component that is represented by any space $X_i$ that is not represented by any compatible system of points. This might make $pi_0 lim X_i to lim pi_0 X_i$ not surjective.
The second is the opposite: the map $pi_0 lim X_i to lim pi_0 X_i$ may not be injective. In this case, you may have two points $x$ and $y$ in the limit such that the images in any individual $x_i$ are connected by a path, but where no path can be compatibly lifted all the way up the tower. For example, if $f:S^1 to S^1$ is a degree-2 covering map, then the limit of the tower $$dots xrightarrow{f} S^1 xrightarrow{f} S^1 xrightarrow{f} S^1$$ is called the 2-adic solenoid and it has uncountably many path components.
The first problem goes away if the maps $X_i to X_{i-1}$ are fibrations, and in this case we often call the limit a homotopy limit. The second problem does not go away in this case, but Milnor proved that $pi_0 lim X_i$ is built out of two terms: $lim(pi_0 X_i)$ and a second term called $lim^1(pi_1 X_i)$. In particular, if the spaces $X_i$ are simply-connected then there are no contributions from the second term, and so there will be an isomorphism $pi_0(lim X_i) to lim pi_0(X_i)$.
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
add a comment |
This is not true, for two distinct reasons.
The first is that the inverse system of spaces may not behave well homotopy-theoretically. If $X_n = [n, infty) subset Bbb R$, then the limit of $dots to X_2 to X_1 to X_0$ is empty. However, on path components it is the constant system $dots to * to * to *$, with limit $*$. Roughly, you might have a path component that is represented by any space $X_i$ that is not represented by any compatible system of points. This might make $pi_0 lim X_i to lim pi_0 X_i$ not surjective.
The second is the opposite: the map $pi_0 lim X_i to lim pi_0 X_i$ may not be injective. In this case, you may have two points $x$ and $y$ in the limit such that the images in any individual $x_i$ are connected by a path, but where no path can be compatibly lifted all the way up the tower. For example, if $f:S^1 to S^1$ is a degree-2 covering map, then the limit of the tower $$dots xrightarrow{f} S^1 xrightarrow{f} S^1 xrightarrow{f} S^1$$ is called the 2-adic solenoid and it has uncountably many path components.
The first problem goes away if the maps $X_i to X_{i-1}$ are fibrations, and in this case we often call the limit a homotopy limit. The second problem does not go away in this case, but Milnor proved that $pi_0 lim X_i$ is built out of two terms: $lim(pi_0 X_i)$ and a second term called $lim^1(pi_1 X_i)$. In particular, if the spaces $X_i$ are simply-connected then there are no contributions from the second term, and so there will be an isomorphism $pi_0(lim X_i) to lim pi_0(X_i)$.
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
This is not true, for two distinct reasons.
The first is that the inverse system of spaces may not behave well homotopy-theoretically. If $X_n = [n, infty) subset Bbb R$, then the limit of $dots to X_2 to X_1 to X_0$ is empty. However, on path components it is the constant system $dots to * to * to *$, with limit $*$. Roughly, you might have a path component that is represented by any space $X_i$ that is not represented by any compatible system of points. This might make $pi_0 lim X_i to lim pi_0 X_i$ not surjective.
The second is the opposite: the map $pi_0 lim X_i to lim pi_0 X_i$ may not be injective. In this case, you may have two points $x$ and $y$ in the limit such that the images in any individual $x_i$ are connected by a path, but where no path can be compatibly lifted all the way up the tower. For example, if $f:S^1 to S^1$ is a degree-2 covering map, then the limit of the tower $$dots xrightarrow{f} S^1 xrightarrow{f} S^1 xrightarrow{f} S^1$$ is called the 2-adic solenoid and it has uncountably many path components.
The first problem goes away if the maps $X_i to X_{i-1}$ are fibrations, and in this case we often call the limit a homotopy limit. The second problem does not go away in this case, but Milnor proved that $pi_0 lim X_i$ is built out of two terms: $lim(pi_0 X_i)$ and a second term called $lim^1(pi_1 X_i)$. In particular, if the spaces $X_i$ are simply-connected then there are no contributions from the second term, and so there will be an isomorphism $pi_0(lim X_i) to lim pi_0(X_i)$.
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
answered Dec 6 at 21:26
Tyler Lawson
38.9k8134198
38.9k8134198
add a comment |
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No. For example, take all your $X_i$ to be the $1$-sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected.
– Achim Krause
Dec 6 at 21:20