Are all maps into path-connected spaces homotopic?
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Context: A problem I'm solving asks "Prove that any two maps $S^m rightarrow S^n$, where $n>m$ is homotopic. [Hint: use the Simplicial Approximation theorem]
My first thoughts were that if we have two maps $f,g$, since $S^n$ is path connected, I can construct a homotopy between any two maps by connecting the two points $f(x)$, $g(x)$ via some path. But the hint makes me think I'm missing something.
algebraic-topology
asked Jul 8 '15 at 4:47
SolveIt
404514
add a comment |Â
up vote
4
down vote
favorite
Context: A problem I'm solving asks "Prove that any two maps $S^m rightarrow S^n$, where $n>m$ is homotopic. [Hint: use the Simplicial Approximation theorem]
My first thoughts were that if we have two maps $f,g$, since $S^n$ is path connected, I can construct a homotopy between any two maps by connecting the two points $f(x)$, $g(x)$ via some path. But the hint makes me think I'm missing something.
algebraic-topology
asked Jul 8 '15 at 4:47
SolveIt
404514
5
You have to use $m>n$. If for example $n=m=1$, then the identity map is not homotopic to a constant map. The paths cannot be chosen continuously.
– user99914
Jul 8 '15 at 4:49
4
The structure of the set $[M,N]$ of homotopy classes of maps $M to N$ is usually very interesting for most $M$ and $N$. It would not be an overstatement to say that understanding this set is one of the primary goals of homotopy theory. (It might be an understatement.)
– Mike Miller
Jul 8 '15 at 4:51
@John Of course. Thank you.
– SolveIt
Jul 8 '15 at 4:54
1
Do you know about CW-complexes? The cellular approximation theorem?
– Arthur
Jul 8 '15 at 5:15
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Context: A problem I'm solving asks "Prove that any two maps $S^m rightarrow S^n$, where $n>m$ is homotopic. [Hint: use the Simplicial Approximation theorem]
My first thoughts were that if we have two maps $f,g$, since $S^n$ is path connected, I can construct a homotopy between any two maps by connecting the two points $f(x)$, $g(x)$ via some path. But the hint makes me think I'm missing something.
algebraic-topology
asked Jul 8 '15 at 4:47
SolveIt
404514
Context: A problem I'm solving asks "Prove that any two maps $S^m rightarrow S^n$, where $n>m$ is homotopic. [Hint: use the Simplicial Approximation theorem]
My first thoughts were that if we have two maps $f,g$, since $S^n$ is path connected, I can construct a homotopy between any two maps by connecting the two points $f(x)$, $g(x)$ via some path. But the hint makes me think I'm missing something.
algebraic-topology
algebraic-topology
asked Jul 8 '15 at 4:47
SolveIt
404514
asked Jul 8 '15 at 4:47
SolveIt
404514
edited Jul 8 '15 at 4:50
user99914
asked Jul 8 '15 at 4:47
SolveIt
404514
asked Jul 8 '15 at 4:47
SolveIt
404514
asked Jul 8 '15 at 4:47
SolveIt
404514
404514
5
You have to use $m>n$. If for example $n=m=1$, then the identity map is not homotopic to a constant map. The paths cannot be chosen continuously.
– user99914
Jul 8 '15 at 4:49
4
The structure of the set $[M,N]$ of homotopy classes of maps $M to N$ is usually very interesting for most $M$ and $N$. It would not be an overstatement to say that understanding this set is one of the primary goals of homotopy theory. (It might be an understatement.)
– Mike Miller
Jul 8 '15 at 4:51
@John Of course. Thank you.
– SolveIt
Jul 8 '15 at 4:54
1
Do you know about CW-complexes? The cellular approximation theorem?
– Arthur
Jul 8 '15 at 5:15
add a comment |Â
5
You have to use $m>n$. If for example $n=m=1$, then the identity map is not homotopic to a constant map. The paths cannot be chosen continuously.
– user99914
Jul 8 '15 at 4:49
4
The structure of the set $[M,N]$ of homotopy classes of maps $M to N$ is usually very interesting for most $M$ and $N$. It would not be an overstatement to say that understanding this set is one of the primary goals of homotopy theory. (It might be an understatement.)
– Mike Miller
Jul 8 '15 at 4:51
@John Of course. Thank you.
– SolveIt
Jul 8 '15 at 4:54
1
Do you know about CW-complexes? The cellular approximation theorem?
– Arthur
Jul 8 '15 at 5:15
5
5
You have to use $m>n$. If for example $n=m=1$, then the identity map is not homotopic to a constant map. The paths cannot be chosen continuously.
– user99914
Jul 8 '15 at 4:49
You have to use $m>n$. If for example $n=m=1$, then the identity map is not homotopic to a constant map. The paths cannot be chosen continuously.
– user99914
Jul 8 '15 at 4:49
4
4
The structure of the set $[M,N]$ of homotopy classes of maps $M to N$ is usually very interesting for most $M$ and $N$. It would not be an overstatement to say that understanding this set is one of the primary goals of homotopy theory. (It might be an understatement.)
– Mike Miller
Jul 8 '15 at 4:51
The structure of the set $[M,N]$ of homotopy classes of maps $M to N$ is usually very interesting for most $M$ and $N$. It would not be an overstatement to say that understanding this set is one of the primary goals of homotopy theory. (It might be an understatement.)
– Mike Miller
Jul 8 '15 at 4:51
@John Of course. Thank you.
– SolveIt
Jul 8 '15 at 4:54
@John Of course. Thank you.
– SolveIt
Jul 8 '15 at 4:54
1
1
Do you know about CW-complexes? The cellular approximation theorem?
– Arthur
Jul 8 '15 at 5:15
Do you know about CW-complexes? The cellular approximation theorem?
– Arthur
Jul 8 '15 at 5:15
add a comment |Â
2 Answers
2
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oldest
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To answer the question in the title: no, of course not, different maps into a path-connected space are not necessarily homotopic. A space $X$ such that for all spaces $Y$, all maps $f,g Y to X$ are homotopic is either contractible or empty. Indeed, if $X$ is nonempty, choose $Y = X$, $f = operatornameid_X$ and $g$ a constant map; then the identity is homotopic to a constant map so by definition $X$ is contractible.
It is true that for all values of $x$, you can construct a path between $f(x)$ and $g(x)$; but it's not necessarily possible to choose such a homotopy continuously. If the map is not nullhomotopic then at some point there will be some kind of tearing, and the resulting map will not be continuous.
To answer the question in the body: there is a simplicial complex structure on $S^k$ with exactly two cells, one in dimension $0$ and one in dimension $k$. So if $f : S^m to S^n$ is a map, $m < n$, by the simplicial approximation theorem $f$ is homotopic to a map $g : S^m to S^n$ such that the image of the $m$-skeleton of $S^m$ is included in the $m$-skeleton of $S^n$. But the $m$-skeleton of $S^m$ is $S^m$, whereas the $m$-skeleton of $S^n$ is a point ($m < n$). It follows that $f$ is homotopic to a constant map.
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
add a comment |Â
up vote
0
down vote
Not necessarily.
Take $f: S^n to S^n$ as identity map
And. $g: S^n to S^n$ as constant map
If you supposition is true then $f$ is homotopic to $g$ as $S^n$ is path connected.
But it is contradicting the fact that $S^n$ is not contractible
edited Sep 3 at 9:26
Jneven
633320
answered Sep 3 at 9:04
Sahil Kalra
1
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
To answer the question in the title: no, of course not, different maps into a path-connected space are not necessarily homotopic. A space $X$ such that for all spaces $Y$, all maps $f,g Y to X$ are homotopic is either contractible or empty. Indeed, if $X$ is nonempty, choose $Y = X$, $f = operatornameid_X$ and $g$ a constant map; then the identity is homotopic to a constant map so by definition $X$ is contractible.
It is true that for all values of $x$, you can construct a path between $f(x)$ and $g(x)$; but it's not necessarily possible to choose such a homotopy continuously. If the map is not nullhomotopic then at some point there will be some kind of tearing, and the resulting map will not be continuous.
To answer the question in the body: there is a simplicial complex structure on $S^k$ with exactly two cells, one in dimension $0$ and one in dimension $k$. So if $f : S^m to S^n$ is a map, $m < n$, by the simplicial approximation theorem $f$ is homotopic to a map $g : S^m to S^n$ such that the image of the $m$-skeleton of $S^m$ is included in the $m$-skeleton of $S^n$. But the $m$-skeleton of $S^m$ is $S^m$, whereas the $m$-skeleton of $S^n$ is a point ($m < n$). It follows that $f$ is homotopic to a constant map.
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
add a comment |Â
up vote
2
down vote
accepted
To answer the question in the title: no, of course not, different maps into a path-connected space are not necessarily homotopic. A space $X$ such that for all spaces $Y$, all maps $f,g Y to X$ are homotopic is either contractible or empty. Indeed, if $X$ is nonempty, choose $Y = X$, $f = operatornameid_X$ and $g$ a constant map; then the identity is homotopic to a constant map so by definition $X$ is contractible.
It is true that for all values of $x$, you can construct a path between $f(x)$ and $g(x)$; but it's not necessarily possible to choose such a homotopy continuously. If the map is not nullhomotopic then at some point there will be some kind of tearing, and the resulting map will not be continuous.
To answer the question in the body: there is a simplicial complex structure on $S^k$ with exactly two cells, one in dimension $0$ and one in dimension $k$. So if $f : S^m to S^n$ is a map, $m < n$, by the simplicial approximation theorem $f$ is homotopic to a map $g : S^m to S^n$ such that the image of the $m$-skeleton of $S^m$ is included in the $m$-skeleton of $S^n$. But the $m$-skeleton of $S^m$ is $S^m$, whereas the $m$-skeleton of $S^n$ is a point ($m < n$). It follows that $f$ is homotopic to a constant map.
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
To answer the question in the title: no, of course not, different maps into a path-connected space are not necessarily homotopic. A space $X$ such that for all spaces $Y$, all maps $f,g Y to X$ are homotopic is either contractible or empty. Indeed, if $X$ is nonempty, choose $Y = X$, $f = operatornameid_X$ and $g$ a constant map; then the identity is homotopic to a constant map so by definition $X$ is contractible.
It is true that for all values of $x$, you can construct a path between $f(x)$ and $g(x)$; but it's not necessarily possible to choose such a homotopy continuously. If the map is not nullhomotopic then at some point there will be some kind of tearing, and the resulting map will not be continuous.
To answer the question in the body: there is a simplicial complex structure on $S^k$ with exactly two cells, one in dimension $0$ and one in dimension $k$. So if $f : S^m to S^n$ is a map, $m < n$, by the simplicial approximation theorem $f$ is homotopic to a map $g : S^m to S^n$ such that the image of the $m$-skeleton of $S^m$ is included in the $m$-skeleton of $S^n$. But the $m$-skeleton of $S^m$ is $S^m$, whereas the $m$-skeleton of $S^n$ is a point ($m < n$). It follows that $f$ is homotopic to a constant map.
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
To answer the question in the title: no, of course not, different maps into a path-connected space are not necessarily homotopic. A space $X$ such that for all spaces $Y$, all maps $f,g Y to X$ are homotopic is either contractible or empty. Indeed, if $X$ is nonempty, choose $Y = X$, $f = operatornameid_X$ and $g$ a constant map; then the identity is homotopic to a constant map so by definition $X$ is contractible.
It is true that for all values of $x$, you can construct a path between $f(x)$ and $g(x)$; but it's not necessarily possible to choose such a homotopy continuously. If the map is not nullhomotopic then at some point there will be some kind of tearing, and the resulting map will not be continuous.
To answer the question in the body: there is a simplicial complex structure on $S^k$ with exactly two cells, one in dimension $0$ and one in dimension $k$. So if $f : S^m to S^n$ is a map, $m < n$, by the simplicial approximation theorem $f$ is homotopic to a map $g : S^m to S^n$ such that the image of the $m$-skeleton of $S^m$ is included in the $m$-skeleton of $S^n$. But the $m$-skeleton of $S^m$ is $S^m$, whereas the $m$-skeleton of $S^n$ is a point ($m < n$). It follows that $f$ is homotopic to a constant map.
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
answered Jul 10 '15 at 7:45
Najib Idrissi
39.4k469135
39.4k469135
add a comment |Â
add a comment |Â
up vote
0
down vote
Not necessarily.
Take $f: S^n to S^n$ as identity map
And. $g: S^n to S^n$ as constant map
If you supposition is true then $f$ is homotopic to $g$ as $S^n$ is path connected.
But it is contradicting the fact that $S^n$ is not contractible
edited Sep 3 at 9:26
Jneven
633320
answered Sep 3 at 9:04
Sahil Kalra
1
add a comment |Â
up vote
0
down vote
Not necessarily.
Take $f: S^n to S^n$ as identity map
And. $g: S^n to S^n$ as constant map
If you supposition is true then $f$ is homotopic to $g$ as $S^n$ is path connected.
But it is contradicting the fact that $S^n$ is not contractible
edited Sep 3 at 9:26
Jneven
633320
answered Sep 3 at 9:04
Sahil Kalra
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Not necessarily.
Take $f: S^n to S^n$ as identity map
And. $g: S^n to S^n$ as constant map
If you supposition is true then $f$ is homotopic to $g$ as $S^n$ is path connected.
But it is contradicting the fact that $S^n$ is not contractible
edited Sep 3 at 9:26
Jneven
633320
answered Sep 3 at 9:04
Sahil Kalra
1
Not necessarily.
Take $f: S^n to S^n$ as identity map
And. $g: S^n to S^n$ as constant map
If you supposition is true then $f$ is homotopic to $g$ as $S^n$ is path connected.
But it is contradicting the fact that $S^n$ is not contractible
edited Sep 3 at 9:26
Jneven
633320
answered Sep 3 at 9:04
Sahil Kalra
1
edited Sep 3 at 9:26
Jneven
633320
edited Sep 3 at 9:26
Jneven
633320
edited Sep 3 at 9:26
Jneven
633320
633320
answered Sep 3 at 9:04
Sahil Kalra
1
answered Sep 3 at 9:04
Sahil Kalra
1
answered Sep 3 at 9:04
Sahil Kalra
1
1
add a comment |Â
add a comment |Â
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5
You have to use $m>n$. If for example $n=m=1$, then the identity map is not homotopic to a constant map. The paths cannot be chosen continuously.
– user99914
Jul 8 '15 at 4:49
4
The structure of the set $[M,N]$ of homotopy classes of maps $M to N$ is usually very interesting for most $M$ and $N$. It would not be an overstatement to say that understanding this set is one of the primary goals of homotopy theory. (It might be an understatement.)
– Mike Miller
Jul 8 '15 at 4:51
@John Of course. Thank you.
– SolveIt
Jul 8 '15 at 4:54
1
Do you know about CW-complexes? The cellular approximation theorem?
– Arthur
Jul 8 '15 at 5:15