Apparent Contradiction: Every Map from a Contractible Space to any X is trivi

[Bacle]

Hi, everyone:

I am confused about the result that every map from a contractible space X into
any topological space Y is contractible.

I think the caveat here is that the homotopy between any f:X-->Y and c:X-->Y

with c(X)={pt.} is that the homotopy is free, i.e., the endpoints are not fixed.


Still, I find the example of f:I-->^{S^1} , with I the unit interval,
particularly confusing:

Consider the path f(t)=^{e^2iPit}. According to this result, since
I is contractible ( I am contractible?:smile) :, f(t) is trivial, i.e., f(t) is homotopic
to a point.

*But* since ^{e^2iPit} does a full loop around ^{S^1} , i.e,
f(I)= (^{S^1}), we must tear f(I) open , to be able to homotope it
to a point. Isn't this tearing necessarily discontinuous, even if we do not fix the
endpoints in ^{S^1}.?.
If not, what would be an actual homotopy between ^{e^2iPit} and {.pt.}?

Thanks.

 

[zhentil]

Hi Bacle. The easy answer is that if a space retracts onto a point, i.e. we have a map

F:X \times I \rightarrow X

such that F restricted to X \times {0} is the identity, and F restricted to X \times {1} is constant, then a homotopy is constructed using f(F).

 

[Hurkyl]

Saying that a map f:X --> Y is homotopic to a constant map is not the same thing as saying that f(X) is contractible.

 

[zhentil]

If that were true, every space would be contractible.

 

[Bacle]

Thanks to Both.

Hurkyl wrote:

" Saying that a map f:X --> Y is homotopic to a constant map is not the same thing as saying that f(X) is contractible. "

I understand, and that is the whole point. I think I know
how to show this formally (see below) , but I cannot see it geometrically/intuitively.
I fail to see intuitively/geometrically how we could deform the circle into a point in a
continuous way.

I had been working along the lines of what Zhentil suggested. I had
constructed a homotopy F(x,t) , from the unit interval I to {1}:

F(x,t)=(1-t)x+t , then H(x,0)=x ; H(x,1)=1 . Using f(t)=e^i2Pit

Then f(F(x,t))=f( (1-t)x+t)= e^{i2Pi((1-t)x+t)}

and f(F(x,0))=e^i2Pix , and f(F(x,1))=e^i2Pi=(1,0)


Which I think is formally correct, tho showing the continuity of f(F) is kind of
a hassle.

What seems confusing is that in this case f(X)=Y . So f(X) coincides with the
whole space Y . How isn't then the contractibility of f(X) not equivalent to
f:X-->Y being inessential.?. If f(x) were not onto, the question would be trivial
to me. What is confusing is that f(X)=Y in this case.

Thanks.

 

[Hurkyl]

You can't deform a circle to a point -- but you can deform a map from the interval to a circle to a map from the interval to a point.

Maybe your problem is that you're confusing the map with its image?

0 and 1 are different points of the interval, so a homotopy g-->h is allowed to "move" the images of those points independently.

But by focusing on the image of f, you've "forgotten" that 0 and 1 are different, which makes it difficult for you to picture how your homotopy works.

 

[Bacle]

Thanks, Hurkyl. Just one more please. (if I don't get it after this one, I will

just meditate on it some more.)


Please tell me if this is correct: this homotopy actually _does_ tear the circle

open. But because of what you said ( 1 and 0 are not close by.) , tearing the

circle (which is, I think, what we do when go from:
:

f(F(x,0))=e^i2Pix, to:

e^{i2Pi((1-eps.)t+eps.} , with eps.->0⁺

is not discontinuous, because it is not the case that points nearby are not sent

(via the tearing) to points "far-away" , i.e., tearing does not violate a del.-eps

argument (which we can use for X,Y metric).

Sorry if this is contrived.

Thanks Again.

 

[Bacle]

Thanks, Hurkyl. Just one more please. (if I don't get it after this one, I will

just meditate on it some more.)


Please tell me if this is correct: this homotopy actually _does_ tear the circle

open. But because of what you said ( 1 and 0 are not close by --as points in I, with
metric |x-y|.) , tearing the

circle (which is, I think, what we do when go from:
:

f(F(x,0))=e^i2Pix, to:

e^{i2Pi((1-eps.)t+eps.} , with eps.->0⁺ )

is not discontinuous, because it is not the case that points nearby are not sent

(via the tearing) to points "far-away" , i.e., tearing does not violate a del.-eps

argument (which we can use for X,Y metric). Points near (1,0) , both from above,

and from below, are points whose preimages are not close-by.

Sorry if this is contrived.

Thanks Again For your Patience.

 

[zhentil]

Think of it this way: take a shoe string, and line up the ends so it forms a circle. You can pull it apart without tearing, i.e. on the interval, 0 and 1 are distinct points, but on the circle, they aren't (if we view the circle as the quotient of the interval).

 

[Hurkyl]

Maybe the words you are using invoke a different mental picture to me than they do to you. It really sounds like the words you use are talking about functions and homotopies from the circle, rather than from the interval.


Now, I could imagine thinking about the following:

We can interpret the inverse image of f:I --> S¹ as a sheaf on S¹ -- over each point of S¹ we have a single point, except for 1, over which we have two points.

There is probably a sense in which we could compose with the homotopy F, giving us a picture of the two points lying over 1 separating; one staying in place and the other moving around the circle.