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homotopy inverse of the cap product
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The cap product with the fundamental cycle of an oriented, triangulated $n$-manifold $X$ defines a chain map $-cap X: C^*(X,A) to C_n-*(X,A)$, where $A$ is any abelian coefficient group.
It's easy to see by computing examples that this map is neither surjective nor injective in general. However, it induces isomorphisms $H^*(X,A) simeq H_*-n(X,A)$ so it is a quasi-isomorphism.
Hence my question: is there any well-known or simple construction of an "inverse map"
$$f:C_*(X,A) to C^*-n(X,A)$$
such that $f(-) cap X$ and $f(- cap X)$ are homotopic to the identity maps on $C_*$ and $C^*$, respectively?
algebraic-topology homology-cohomology
asked Jul 23 at 0:23
wzzx
270111
add a comment |Â
up vote
3
down vote
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The cap product with the fundamental cycle of an oriented, triangulated $n$-manifold $X$ defines a chain map $-cap X: C^*(X,A) to C_n-*(X,A)$, where $A$ is any abelian coefficient group.
It's easy to see by computing examples that this map is neither surjective nor injective in general. However, it induces isomorphisms $H^*(X,A) simeq H_*-n(X,A)$ so it is a quasi-isomorphism.
Hence my question: is there any well-known or simple construction of an "inverse map"
$$f:C_*(X,A) to C^*-n(X,A)$$
such that $f(-) cap X$ and $f(- cap X)$ are homotopic to the identity maps on $C_*$ and $C^*$, respectively?
algebraic-topology homology-cohomology
asked Jul 23 at 0:23
wzzx
270111
You need $X$ compact (or to take Borel-Moore homology, or compactly supported cohomology) if you want such an isomorphism.
– Nicolas Hemelsoet
Jul 23 at 9:39
Thanks, although the maps are (locally) indistinguishable in all those cases, they just induce isomorphisms on different versions of the cohomology of the complex.
– wzzx
Jul 23 at 19:37
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The cap product with the fundamental cycle of an oriented, triangulated $n$-manifold $X$ defines a chain map $-cap X: C^*(X,A) to C_n-*(X,A)$, where $A$ is any abelian coefficient group.
It's easy to see by computing examples that this map is neither surjective nor injective in general. However, it induces isomorphisms $H^*(X,A) simeq H_*-n(X,A)$ so it is a quasi-isomorphism.
Hence my question: is there any well-known or simple construction of an "inverse map"
$$f:C_*(X,A) to C^*-n(X,A)$$
such that $f(-) cap X$ and $f(- cap X)$ are homotopic to the identity maps on $C_*$ and $C^*$, respectively?
algebraic-topology homology-cohomology
asked Jul 23 at 0:23
wzzx
270111
The cap product with the fundamental cycle of an oriented, triangulated $n$-manifold $X$ defines a chain map $-cap X: C^*(X,A) to C_n-*(X,A)$, where $A$ is any abelian coefficient group.
It's easy to see by computing examples that this map is neither surjective nor injective in general. However, it induces isomorphisms $H^*(X,A) simeq H_*-n(X,A)$ so it is a quasi-isomorphism.
Hence my question: is there any well-known or simple construction of an "inverse map"
$$f:C_*(X,A) to C^*-n(X,A)$$
such that $f(-) cap X$ and $f(- cap X)$ are homotopic to the identity maps on $C_*$ and $C^*$, respectively?
algebraic-topology homology-cohomology
asked Jul 23 at 0:23
wzzx
270111
asked Jul 23 at 0:23
wzzx
270111
asked Jul 23 at 0:23
wzzx
270111
asked Jul 23 at 0:23
wzzx
270111
270111
You need $X$ compact (or to take Borel-Moore homology, or compactly supported cohomology) if you want such an isomorphism.
– Nicolas Hemelsoet
Jul 23 at 9:39
Thanks, although the maps are (locally) indistinguishable in all those cases, they just induce isomorphisms on different versions of the cohomology of the complex.
– wzzx
Jul 23 at 19:37
add a comment |Â
You need $X$ compact (or to take Borel-Moore homology, or compactly supported cohomology) if you want such an isomorphism.
– Nicolas Hemelsoet
Jul 23 at 9:39
Thanks, although the maps are (locally) indistinguishable in all those cases, they just induce isomorphisms on different versions of the cohomology of the complex.
– wzzx
Jul 23 at 19:37
You need $X$ compact (or to take Borel-Moore homology, or compactly supported cohomology) if you want such an isomorphism.
– Nicolas Hemelsoet
Jul 23 at 9:39
You need $X$ compact (or to take Borel-Moore homology, or compactly supported cohomology) if you want such an isomorphism.
– Nicolas Hemelsoet
Jul 23 at 9:39
Thanks, although the maps are (locally) indistinguishable in all those cases, they just induce isomorphisms on different versions of the cohomology of the complex.
– wzzx
Jul 23 at 19:37
Thanks, although the maps are (locally) indistinguishable in all those cases, they just induce isomorphisms on different versions of the cohomology of the complex.
– wzzx
Jul 23 at 19:37
add a comment |Â
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You need $X$ compact (or to take Borel-Moore homology, or compactly supported cohomology) if you want such an isomorphism.
– Nicolas Hemelsoet
Jul 23 at 9:39
Thanks, although the maps are (locally) indistinguishable in all those cases, they just induce isomorphisms on different versions of the cohomology of the complex.
– wzzx
Jul 23 at 19:37