Contractible Image.

[hbe...@gmail.com]

Hi, All:

Let f: X-->Y be an innesential map, i.e., the image f(X) is not continuously
contractible to a point. Does it follow that the induced map on homology is non-trivial (and, does the converse hold, i.e., does any innessential map induce
a trivial ( i.e., zero) map in homology? I imagine any contractible subspace S has trivial homology, and trivial homotopy ; at least Pi_1(S)=0 .
Thanks.

[W. Dale Hall]

A spelling correction: the word is inessential. It means the opposite
of what you wrote. An inessential map is (in the context of homotopy
theory) a null-homotopic map [i.e., continuously deformable to a point].
The other maps are essential.

Now, for your question: an essential map does not necessarily produce
nonzero homomorphisms in homology. The easiest example I come up with
is the Hopf map h : S^3 --> S^2

(The MathWorld page http://mathworld.wolfram.com/HopfMap.html may be
more informative than what follows)

Let S^3 denote the set of points

S^3 = {(z1, z2) in C^2 | |z1|^2 + |z2|^2 = 1}

The sphere S^1 = {z in C | |z| = 1} acts freely on S^3

z * (z1,z2) = (z*z1, z*z2)

with orbit space equal to the complex projective line

CP^1 = S^3/{(y,z) ~ t*(y,z) for t in C\{0} }

but CP^1 is S^2, so h: S^3 --> S^2

To see that the induced homomorphisms on (reduced) homology
are trivial, note that reduced homology for S^3 is nonzero
only in dimension 3, while reduced homology for S^2 is
nonzero only in dimension 2. With nowhere to live and nowhere
to go to, the induced homomorphism dies.

To see that the map h is not null-homotopic, note that the
fibres of h (the preimages of distinct points of S^2) are
linked; their linking number is called the Hopf invariant,
and for this map, that invariant is 1. Proving that the
Hopf invariant is invariant under homotopy, and that the
invariant of a null-homotopic map would be zero then completes
the proof.

[hbe...@gmail.com]

On Tuesday, May 21, 2013 2:21:01 PM UTC-4, W. Dale Hall wrote:

Thanks, W.Dale Hall. But I think I asked the wrong question ( or maybe I

misunderstood--or misunderestimated-- your reply; sorry if this is so), or at

least I did not ask the 2nd half of an iff question. Would you please

help me with the following : so, we have an innesential map f:X-->Y

(correcting the use of essential instead of innessential). Can f(X)

represent a non-zero homology class? I know that there are relations between

homotopy and homology , e.g., Hurewicz theorem, but I don't know if f(X)

being homotopically-trivial in Y implies that f(X) is also homologically-

trivial. Morover, can any non-trivial homology class in a general space

be represented by a contractible subspace?

Thanks

[W. Dale Hall]

hbe...@gmail.com wrote:
> On Tuesday, May 21, 2013 2:21:01 PM UTC-4, W. Dale Hall wrote:
>> hbail.com wrote:
>>
>>> Hi, All:
>>
>>>
>>
>>> Let f: X-->Y be an innesential map, i.e., the image f(X) is not
>>
>>> continuously contractible to a point. Does it follow that the induced
>>
>>> map on homology is non-trivial (and, does the converse hold, i.e.,
>>
>>> does any innessential map induce a trivial ( i.e., zero) map in
>>
>>> homology? I imagine any contractible subspace S has trivial homology,
>>
>>> and trivial homotopy ; at least Pi_1(S)=0 . Thanks.
>>
>>>

... irrelevant response deleted

>
> Thanks, W.Dale Hall. But I think I asked the wrong question ( or maybe I
>
> misunderstood--or misunderestimated-- your reply; sorry if this is so), or at
>
> least I did not ask the 2nd half of an iff question. Would you please
>
> help me with the following : so, we have an innesential map f:X-->Y
>
> (correcting the use of essential instead of innessential). Can f(X)
>
> represent a non-zero homology class? I know that there are relations between
>
> homotopy and homology , e.g., Hurewicz theorem, but I don't know if f(X)
>
> being homotopically-trivial in Y implies that f(X) is also homologically-
>
> trivial. Morover, can any non-trivial homology class in a general space
>
> be represented by a contractible subspace?
>
> Thanks
>

(a minor nit .. inessential: one N and two S's)

A null-homotopic map always induces a zero homomorphism in reduced
homology. To see this, note if f1, f2 are homotopic maps from X to Y,
then the induced homomorphisms f1* and f2* are identical. In simplicial
or singular homology, a homotopy F : X x I ---> Y is used to produce
a chain homotopy between f1* and f2* (at the level of chains), and in
passing from chains to homology, the two homomorphisms become equal.

Having done that, replace your null-homotopic map f by a constant map k,
and note that the image of k* is a subset of H*(point) --> H*(Y), where
the arrow is induced by inclusion.

Dale

[bacle...@gmail.com]

I think you can use the degree of the map, which is a homotopy-invariant.

The degree of a contractible map is 0.