Simple Question on Homotopy Equiv. and Contractible Subspaces

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Hi, everyone:

I have not found a clear explanation for this:

Are any two contractible subspaces A,B<X homotopy-equivalent to each other?

Clearly, homotopy equivalence is an equivalence relation. BUT: what if

A<X is H-E (Homotopy-Equiv.) to a point p in X, while B is H-E to a point q

p=/q ? (like in the case where X is disconnected, and A,B are in different

components of X ) . If we had p=q, then the answer is trivially yes; we compose

a homotopy of A with p , with the reverse homotopy of B with p.


Also: when we say that the identity map id:X-->X from any topological space

to itself is not nullhomotopic, i.e., homotopic to a point. Is this always equivalent

to saying that X is not contractible? I have a proof saying yes, but one part

of the proof is kind of shaky.

Hope this is not too dumb.
 

Answers and Replies

  • #2
quasar987
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1) If p=/q, let f:{p}-->{q} be the map f(p)=q (a homeomorphism) and consider instead the composition of "a homotopy of A with p" o f o "the reverse homotopy of B with q".

2) To say that the identity map id:X-->X is nullhomotopic (i.e homotopy to a constant map r:x-->p [which can also be seen as a retraction]) is equivalent to saying that the retraction r:X-->{p} and the inclusion i:{p}-->X are homotopy inverses of each other.
So id nullhomotopic is the same as X homotopy equivalent to a point, i.e. contractible.
 
  • #3
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Thanks, Q, but this is specifically what I am confused with:

Say we have a subspace D of R^2 consisting of two disjoint disks D1,D2 that do not
share a limit point , i.e., the subpace D is then disconnected.

Then D1 is homotopic to a point p in D1, D2 is homotopic to a point q in D2.
And p is homotopic to q .

We then have just one homotopy class in D; that of a point. But if the only
homotopy class of spaces in D is that of a point, then D is contractible.
But D is not path-connected, since it is not connected. And contractible
spaces are path-connected.

So , what is wrong with the argument?
 
  • #4
quasar987
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"We then have just one homotopy class in D"

What do you mean by that? what is "a homotopy class in D"?

(Note that D does not have the homotopy type of a point: it has the same homotopy type as the discrete space consisting of 2 points).
 
  • #5
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I think the problem is my own confusion; I don't have a good feel for the
concept of homotopy equivalence; I do understand the formalities. of having
f:X-->Y, g:Y-->X , with either fog~IdY , or gof~idX, or both

I think this is the issue: usually, if/when we say that a space X is contractible,
then it is homotopic to a point, or that X has the homotopy type of a point,
and the contraction must happen within X itself. And I am having some trouble
wrapping my head around the meaning of a map being homotopic to the identity.

It then seems that X being homotopic to Y is not related to Y being continuously
deformable into X , since otherwise, subspaces in different components could not be
homotopic to each other.

Thanks for any Comments.
 
  • #6
quasar987
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Note first that we do not speak of two spaces X, Y being homotopic. We say that they are of the same homotopy type. And this means that there are continuous maps f:X-->Y, g:Y-->X , with fog~IdY and gof~idX. (not either, but both!)

That is, the compositions fog and gof are both homotopic to the identity. And a map F:X-->X is homotopic to the identity if there is a continuous map H(t,x) such that H(0,x)=IdX and H(1,x)=F. So visually: F has some effect on the points in X, and F~IdX iff this effect can be realized by sliding continuously the points of X around as if following the flow of some liquid on X.

Now, the concept of two spaces being of the same homotopy type is a generalization of the concept of homeomorphism: for X,Y to be homeomorphic, we ask for maps f:X-->Y, g:Y-->X , with fog=IdY and gof=idX (i.e., as you apply g followed by f on the points of Y, you end up on the same points as you started with, and similarly for gof). For same homotopy type, we only ask that as you apply g followed by f on the points of Y, you end up not necessarily on the same points as you started with, but on on a configuration that can be slid around to them in a countinuous manner as if following the flow of some liquid on Y.

And similarly for gof.
 
  • #7
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Thanks, Quasar, and sorry for the somewhat-confused post; I will hopefully get
some sleep soon.

I was just looking for spaces X,Y, so that there are f,g with fog~IdX, but there
are no h,j with hoj~IdY. Isn't this possible? I don't see how the existence of f,g
with fog~IdX implies that gof~IdY

Also, I do get that h.equiv. is a weakening of homeomorphism, but, unless the
homotopy happens within a familiar space like R^n, I don't get the deformation
thing very clearly.
 
  • #8
quasar987
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Thanks, Quasar, and sorry for the somewhat-confused post; I will hopefully get
some sleep soon.

I was just looking for spaces X,Y, so that there are f,g with fog~IdX, but there
are no h,j with hoj~IdY. Isn't this possible? I don't see how the existence of f,g
with fog~IdX implies that gof~IdY
X=S^1, Y={p:=(1,0)}, f:S^1-->{p}, g:{p}-->S^1 the inclusion. fog=idY but gof is not homotopic to the identity. (This we know, say, because pi_1(S^1) = Z while pi_1({p})=0.)

Also, I do get that h.equiv. is a weakening of homeomorphism, but, unless the
homotopy happens within a familiar space like R^n, I don't get the deformation
thing very clearly.
Well, me neither and it doesn't matter, it's just a picture to get some intuitive feel for what same homotopy type means.
 
  • #9
lavinia
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Also, I do get that h.equiv. is a weakening of homeomorphism, but, unless the
homotopy happens within a familiar space like R^n, I don't get the deformation
thing very clearly.
Whenever one space is a strong deformation retraction of the other then they are homotopically equivalent. The examples abound and are easy to imagine.

Take any compact manifold inside another a think of the solid tube surrounding it. For instance in R^3 the tube around a circle is a solid torus. The tube around the equator of the earth is a curved cylinder. the tube around the equator of the projective plane is a Mobius band. Any of these tubes can be shrunk onto the manifold by a homotopy that just slides each point to the nearest point on the manifold. These are all homotopy equivalences.

Here is different one. take R^3 minus the z-axis and the circle of radius one in the xy-plane.

Draw a torus around this circle that doesn't intersect the z-axis. This you can do by giving it a small radius.

Slide each point in R^3 minus the z-axis and the circle to its nearest point on this torus.
This also proves that this space has an abelian fundamental group.
 
  • #10
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Just this last one, please, Quasar:

Are you using, to conclude that gof not ~ IdY (sorry; for some reason, the quoting
function is not enabled in my browser) that :

gof~idY , then (gof)_*=g_*o f_*= IdY_* ?
 
  • #11
quasar987
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The argument I based "gof not ~ idX" on is that if gof was homotopic to idX, then it would be what X=S^1 and Y={p} are of the same homotopy type, and so, in particular, the first homotopy group of X=S^1 and Y={p} would be isomorphic, which is not the case since pi_1(S^1)=Z while pi_1({p})=0.
 
  • #12
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Just to say thanks for everything, Quasar, very helpful--as usual.
 
  • #13
quasar987
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You're welcomed Bacle, I enjoy your questions.
 

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