# Simple Question on Homotopy Equiv. and Contractible Subspaces

## Main Question or Discussion Point

Hi, everyone:

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.

Related Differential Geometry News on Phys.org
quasar987
Homework Helper
Gold Member
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.

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?

quasar987
Homework Helper
Gold Member
"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).

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.

quasar987
Homework Helper
Gold Member
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.

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.

quasar987
Homework Helper
Gold Member
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.

lavinia
Gold Member
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.

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_* ?

quasar987
Homework Helper
Gold Member
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.

Just to say thanks for everything, Quasar, very helpful--as usual.

quasar987