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From: adamk on 2 Aug 2010 15:53
Hi:
Hope this is not too simple-minded:

Given a map f between Chain Complexes C_n(X) and C_n(Y);
X,Y topological spaces, we have that f passes to
homology if f maps cycles into cycles and boundaries
into boundaries. Then we get an induced map on homology,
i.e., a map f_* from H_n(X) into H_n(Y).

Is this condition sufficient(necessary.?) for maps
to pass to cohomology, using Cochain Complexes, i.e.,
the map f sends cocycles to cocycles and coboundaries to
coboundaries, then we get a map f* from H^n(X) to
H^n(Y).?

Thanks.
From: W. Dale Hall on 3 Aug 2010 06:14
adamk wrote:
> Hi:
> Hope this is not too simple-minded:
>
> Given a map f between Chain Complexes C_n(X) and C_n(Y);
> X,Y topological spaces, we have that f passes to
> homology if f maps cycles into cycles and boundaries
> into boundaries. Then we get an induced map on homology,
> i.e., a map f_* from H_n(X) into H_n(Y).
>
> Is this condition sufficient(necessary.?) for maps
> to pass to cohomology, using Cochain Complexes, i.e.,
> the map f sends cocycles to cocycles and coboundaries to
> coboundaries, then we get a map f* from H^n(X) to
> H^n(Y).?
>
> Thanks.

You need a chain map. That's all.

A chain map is (as luck would have it) a mapping between
chain complexes; as such, it comprises a homomorphism
at each dimension

f_n : C_n(X) ---> C_n(Y)

defined in such a way that this diagram

f_n
C_n(X) ------> C_n(Y)

| |
d | | d
| |
V V

C_(n+1)(X) ---> C_(n+1)(Y)
f_(n+1)

commutes, with d being the differential of each
appropriate chain complex.

It is a standard result of homological algebra that
each such chain map induces homomorphisms in both
homology and cohomology. I'd be astonished if any
introductory text on the topic failed to give a
proof.

Dale
From: adamk on 3 Aug 2010 14:11
Thanks, Mr. Hall.

Only result I am aware of ( I have looked at both
Munkres' A.T, and C.R.F Maunder's Algebraic Topology)
re induced maps, is that these maps result from the
following:

Given groups G,G', and respective Normal subgroups
N<G, and N'<G' , if we have a homomorphism h:G-->G'
with h(N)<< N' (with << set containment), then we
get a map f_* : G/N -->G'/N', with f_*([g]):=([f(g)])

This map is the only one ( universal property-wise)
that makes the diagram:

h
G->G'
| |
G/N-->G'/N'
h_*

And the side maps are the natural projections.

In homology, we deal with Abelian groups, so that
any subgroup of C_n(X) is normal in the chain
group C_n(X).

Then, in particular, given a map f:X-->Y
(X,Y top. spaces) ,if (the normal subgroup
of C_n(X) given by) cycles are sent to cycles
and (the normal....of) boundaries are sent to
boundaries, then , from f:X-->Y, we get a map:


f_*: Z_n(X)/B_n(X)--->Z_n(Y)/B_n(Y)


for each n, with f_* defined as above,


And this is what is meant by f "passing to
homology".

But I have not seen any similar result for
cohomology.

Thanks for your comments, though.
From: W. Dale Hall on 4 Aug 2010 03:48
adamk wrote:
> Thanks, Mr. Hall.
>
> Only result I am aware of ( I have looked at both
> Munkres' A.T, and C.R.F Maunder's Algebraic Topology)
> re induced maps, is that these maps result from the
> following:
>
> Given groups G,G', and respective Normal subgroups
> N<G, and N'<G' , if we have a homomorphism h:G-->G'
> with h(N)<< N' (with<< set containment), then we
> get a map f_* : G/N -->G'/N', with f_*([g]):=([f(g)])
>
> This map is the only one ( universal property-wise)
> that makes the diagram:
>
> h
> G->G'
> | |
> G/N-->G'/N'
> h_*
>
> And the side maps are the natural projections.
>
> In homology, we deal with Abelian groups, so that
> any subgroup of C_n(X) is normal in the chain
> group C_n(X).
>
> Then, in particular, given a map f:X-->Y
> (X,Y top. spaces) ,if (the normal subgroup
> of C_n(X) given by) cycles are sent to cycles
> and (the normal....of) boundaries are sent to
> boundaries, then , from f:X-->Y, we get a map:
>
>
> f_*: Z_n(X)/B_n(X)--->Z_n(Y)/B_n(Y)
>
>
> for each n, with f_* defined as above,

For f: X ---> Y, the induced homomorphism on
chain complexes

C(f) : C(X) --> C(Y)

is in fact a chain map, as I described in my
note. That is, C(f) commutes with the boundary
operator, and so maps boundaries to boundaries
and cycles to cycles. It will then induce a
homomorphism in homology.

For an abelian group (or, commonly, a commutative
ring R) The cochain complexes C^*(X;G) and C^*(Y;G)
(respectively C^*(X;R), C^*(Y;R)) are similarly chain
complexes, defined by

C^*(X;G) = Hom(C(X),G)
C^*(Y;G) = Hom(C(Y),G)

(respectively, Hom_Z(C(X),R), Hom_Z(C(Y),R)), but
with the boundary (called coboundary here)
now raising dimension by 1. The definition of the
coboundary is
d
Hom(C_n(X),G) ---> Hom(C_(n+1)(X),G)

a |--------> ao@

where I have used @ to denote the boundary operator
passing from C_(n+1)(X) to C_n(X). One obtains the
cochain homomorphism from a map f:X ---> Y similarly,
by composing the chain homomorphism with the given
cochain:

a in Hom(C_n(Y),G),

C(f) : C_n(X) ---> C_n(Y)

so if a is in C^n(Y), we write

a
C_n(Y) -------> G

and compose with C(f) as follows:

C(f) a
C_n(X) -------> C_n(Y) -----> G
\ _
\_______________________/|
C^*(f) a

to get an element of C^n(X).

Note that coboundaries are the image of the
coboundary homomorphism of C^*(X),C^*(Y), and
as such, are formed by the composition of elements
of C^*(X), C^*(Y) with the boundary operators on
C(X),C(Y). Since C(f) commutes with the boundary
operator:

C(f) @_X = @_Y C(f)

it automatically happens that C^*(f) commutes with
the coboundary operator d:

C^*(f) @_Y = @_X C^*(f)

This again makes C^*(f) a chain map (now called a
cochain map, since the coboundary raises dimension).

>
>
> And this is what is meant by f "passing to
> homology".
>

And it is what I meant as well, except I restricted
the discussion to homology. What I mentioned was that
it was necessary for the induced homomorphism on chain
groups to respect the boundary operators (which it does),
and (oops, perhaps I neglected to point this out:) that
this forces cycles (the kernel of the boundary operator)
and boundaries (the image of the boundary operator) to
be separately respected by the chain map. The proof is
by simple diagram-chasing.

> But I have not seen any similar result for
> cohomology.
>
> Thanks for your comments, though.

I'd still be really surprised if neither Munkres nor
Maunder made this explicit. Allen Hatcher shows this
on p201 of his text ... here's chapter 3, where
it all takes place:

(http://www.math.cornell.edu/~hatcher/AT/ATch3.pdf)

it's on p17 of the 76-page .pdf file, in the section
titled "Induced Homomorphisms".

I've had a quick look at what I can find via Google:

I've had a look at Munkres's text "Topology (2nd edition)"
online, and if that's your text, it apparently doesn't
touch cohomology, so maybe I got the wrong book. I'll
keep looking.

Maunder's "Algebraic Topology" proves on p110 that a
continuous map of pairs induces a chain map on the
chain complexes, and then proves on p162 that a chain
map induces a chain map (in the opposite direction) on
the corresponding cochain complexes.

I'm guessing you didn't see that, or you're looking
at a different text, or maybe you wanted something
more explicit.

Dale
From: adamk on 4 Aug 2010 15:23
Just to say thanks, Mr. Hall, very cler, very thorough.
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