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(algebraic topology) question about the cellular approximation theorem

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(algebraic topology) question about the cellular approximation theorem

Some questions about cellular homology and cohomologyDecoupling the algebra from the topology in cellular homologyUnderstanding example 2.36 in Hatcher's Algebraic Topology (calculating homology groups)For a CW-complex $X$, $(X, X^{(n)})$ is n-connected (without using cellular approximation)The set $pi_n(X)$ of homotopy classes is singleton if $X$ is given the discrete topologyHow to apply the cellular boundary formula?Simplicial Approximation TheoremIdea behind the proof of Whitehead's Theorem and Compression LemmaQuestion about theorem in Hatcher's Algebraic TopologyCellular approximation theorem

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4

$begingroup$

I have a question about the proof $pi _n(S^k) = 0 ~ for ~ n<k~$ by using cellular approximation theorem. According to the proof, (Wikipedia)

$mathbf{proof} $. Give $S^n, S^k $ their canonical CW-structure, with one 0-cell each, and with one n-cell for $S^n$ and k-cell for $S^k$. Any base-point preserving map $ f: S^n to S^k $ is by the cellular approximation theorem homotopic to a constant map, whence $pi_n(S^k)=0 $

Then, I cannot understand why f is homotopic to a constant map by the cellular approximation theorem.

algebraic-topology homotopy-theory

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asked 9 hours ago

Hanwoong ChoHanwoong Cho

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4

$begingroup$

I have a question about the proof $pi _n(S^k) = 0 ~ for ~ n<k~$ by using cellular approximation theorem. According to the proof, (Wikipedia)

$mathbf{proof} $. Give $S^n, S^k $ their canonical CW-structure, with one 0-cell each, and with one n-cell for $S^n$ and k-cell for $S^k$. Any base-point preserving map $ f: S^n to S^k $ is by the cellular approximation theorem homotopic to a constant map, whence $pi_n(S^k)=0 $

Then, I cannot understand why f is homotopic to a constant map by the cellular approximation theorem.

algebraic-topology homotopy-theory

share|cite|improve this question

asked 9 hours ago

Hanwoong ChoHanwoong Cho

2111 bronze badge

New contributor

Hanwoong Cho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I have a question about the proof $pi _n(S^k) = 0 ~ for ~ n<k~$ by using cellular approximation theorem. According to the proof, (Wikipedia)

$mathbf{proof} $. Give $S^n, S^k $ their canonical CW-structure, with one 0-cell each, and with one n-cell for $S^n$ and k-cell for $S^k$. Any base-point preserving map $ f: S^n to S^k $ is by the cellular approximation theorem homotopic to a constant map, whence $pi_n(S^k)=0 $

Then, I cannot understand why f is homotopic to a constant map by the cellular approximation theorem.

algebraic-topology homotopy-theory

share|cite|improve this question

asked 9 hours ago

Hanwoong ChoHanwoong Cho

2111 bronze badge

New contributor

Hanwoong Cho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 9 hours ago

Hanwoong ChoHanwoong Cho

2111 bronze badge

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Hanwoong Cho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 9 hours ago

Hanwoong ChoHanwoong Cho

2111 bronze badge

New contributor

Hanwoong Cho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

Hanwoong ChoHanwoong Cho

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New contributor

Hanwoong Cho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

Hanwoong ChoHanwoong Cho

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2 Answers
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5

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Since $n<k$, $n$-th skeleton of $S^k$ is just a point, but the $n$-th skeleton of $S^n$ is already $S^n$ itself. Cellular approximation implies that $f$ is homotopic to $g: S^n to S^k$ with the property $$g(S^n)=g(text{sk}^n(S^n)) subset text{sk}^n(S^k)={text{pt} },$$ so $g$ is a map to a point.

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answered 9 hours ago

BananeenBananeen

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Since $f$ is homotopic to some cellular map $g$ we only need to consider $g$ for now. We have that $g$ maps the $j$-skeleton of $S^n$, $0 leq j leq n$, to the $j$-skeleton of $S^k$. As $n < k$ (and we have these CW-structures), you can only map to the point that gives the $0$-skeleton of $S^k$ as we first add new cells at level $k$.

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answered 9 hours ago

ThorWittichThorWittich

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5

$begingroup$

Since $n<k$, $n$-th skeleton of $S^k$ is just a point, but the $n$-th skeleton of $S^n$ is already $S^n$ itself. Cellular approximation implies that $f$ is homotopic to $g: S^n to S^k$ with the property $$g(S^n)=g(text{sk}^n(S^n)) subset text{sk}^n(S^k)={text{pt} },$$ so $g$ is a map to a point.

share|cite|improve this answer

answered 9 hours ago

BananeenBananeen

87366 silver badges1313 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Since $n<k$, $n$-th skeleton of $S^k$ is just a point, but the $n$-th skeleton of $S^n$ is already $S^n$ itself. Cellular approximation implies that $f$ is homotopic to $g: S^n to S^k$ with the property $$g(S^n)=g(text{sk}^n(S^n)) subset text{sk}^n(S^k)={text{pt} },$$ so $g$ is a map to a point.

share|cite|improve this answer

answered 9 hours ago

BananeenBananeen

87366 silver badges1313 bronze badges

$endgroup$

add a comment | 

$begingroup$

Since $n<k$, $n$-th skeleton of $S^k$ is just a point, but the $n$-th skeleton of $S^n$ is already $S^n$ itself. Cellular approximation implies that $f$ is homotopic to $g: S^n to S^k$ with the property $$g(S^n)=g(text{sk}^n(S^n)) subset text{sk}^n(S^k)={text{pt} },$$ so $g$ is a map to a point.

share|cite|improve this answer

answered 9 hours ago

BananeenBananeen

87366 silver badges1313 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

BananeenBananeen

87366 silver badges1313 bronze badges

answered 9 hours ago

BananeenBananeen

87366 silver badges1313 bronze badges

answered 9 hours ago

BananeenBananeen

87366 silver badges1313 bronze badges

1

$begingroup$

Since $f$ is homotopic to some cellular map $g$ we only need to consider $g$ for now. We have that $g$ maps the $j$-skeleton of $S^n$, $0 leq j leq n$, to the $j$-skeleton of $S^k$. As $n < k$ (and we have these CW-structures), you can only map to the point that gives the $0$-skeleton of $S^k$ as we first add new cells at level $k$.

share|cite|improve this answer

answered 9 hours ago

ThorWittichThorWittich

3,25033 silver badges2020 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Since $f$ is homotopic to some cellular map $g$ we only need to consider $g$ for now. We have that $g$ maps the $j$-skeleton of $S^n$, $0 leq j leq n$, to the $j$-skeleton of $S^k$. As $n < k$ (and we have these CW-structures), you can only map to the point that gives the $0$-skeleton of $S^k$ as we first add new cells at level $k$.

share|cite|improve this answer

answered 9 hours ago

ThorWittichThorWittich

3,25033 silver badges2020 bronze badges

$endgroup$

add a comment | 

$begingroup$

Since $f$ is homotopic to some cellular map $g$ we only need to consider $g$ for now. We have that $g$ maps the $j$-skeleton of $S^n$, $0 leq j leq n$, to the $j$-skeleton of $S^k$. As $n < k$ (and we have these CW-structures), you can only map to the point that gives the $0$-skeleton of $S^k$ as we first add new cells at level $k$.

share|cite|improve this answer

answered 9 hours ago

ThorWittichThorWittich

3,25033 silver badges2020 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

ThorWittichThorWittich

3,25033 silver badges2020 bronze badges

answered 9 hours ago

ThorWittichThorWittich

3,25033 silver badges2020 bronze badges

answered 9 hours ago

ThorWittichThorWittich

3,25033 silver badges2020 bronze badges