Skip to main content
added 256 characters in body
Source Link
Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories with such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.

EDIT: I forgot that the question was asking specifically about relative homotopy and homology. Luckily the correspondence outlined above sends (up to a shift in degree) relative homology to relative homotopy, so everything I said above is still true.

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories with such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories with such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.

EDIT: I forgot that the question was asking specifically about relative homotopy and homology. Luckily the correspondence outlined above sends (up to a shift in degree) relative homology to relative homotopy, so everything I said above is still true.

correct stupid grammar mistake.
Source Link
Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories bywith such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories by such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories with such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.

Source Link
Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of spaces as Kan complexes, but think of it as "free $\mathbb{Z}$-module over $X$") such that $\pi_*(\mathbb{Z}[X])=H_*(X)$.

As it turns out, this functor is 1-excisive so if you have an homotopy pushout square applying $\mathbb{Z}[-]$ turns it into an homotopy pullback square. So in fact the observation about homology and homotopy pushouts is an immediate consequence of the homotopy excisivity of homology (basically the excision theorem) and the observation about homotopy and homotopy pullbacks.

In fact this fact is not special to singular homology: any homology theory $E$ has an 1-excisive functor (usually denoted by $\Omega^\infty(E\wedge \Sigma^\infty_+-)$ for reasons that are related to the representability of homology theory by spectra) playing essentially the same role. In fact you can identify homology theories by such functors (this is well covered, I believe, by Goodwillie's Calculus I paper that you are reading). This is a consequence of Brown representability.