Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Fundamental class
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Poincaré duality== {{Main|Poincaré duality}} {{Expand section|date=December 2008}} The Poincaré duality theorem relates the homology and cohomology groups of ''n''-dimensional oriented closed manifolds: if ''R'' is a [[commutative ring]] and ''M'' is an ''n''-dimensional ''R''-orientable closed manifold with fundamental class ''[M]'', then for all ''k'', the map : <math> H^k(M;R) \to H_{n-k}(M;R) </math> given by : <math> \alpha \mapsto [M] \frown \alpha </math> is an isomorphism.<ref name=":0" /> Using the notion of fundamental class for manifolds with boundary, we can extend Poincaré duality to that case too (see [[Lefschetz duality]]). In fact, the [[cap product]] with a fundamental class gives a stronger duality result saying that we have isomorphisms <math>H^q(M, A;R) \cong H_{n-q}(M, B;R)</math>, assuming we have that <math>A, B</math> are <math>(n-1)</math>-dimensional manifolds with <math>\partial A=\partial B= A\cap B</math> and <math>\partial M=A\cup B</math>.<ref name=":0">{{Cite book |last=Hatcher |first=Allen |url=https://pi.math.cornell.edu/~hatcher/AT/ATpage.html |title=Algebraic Topology |date=2002 |publisher=[[Cambridge University Press]] |isbn=9780521795401 |edition=1st |location=Cambridge |page= |pages=241–254 |language=English |mr=1867354 |authorlink=Allen Hatcher}}</ref> See also [[Twisted Poincaré duality]]
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)