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2.CY denotes the cone on Y.Show that any two maps f,g:X->CY are homotopic.

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- Thread starter kakarotyjn
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- #1

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2.CY denotes the cone on Y.Show that any two maps f,g:X->CY are homotopic.

- #2

quasar987

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1. Define a homotopy from f to the identity like so: draw a line from f(x) to x. This line lies inside the disk the boundary of which is C, so it's not an homotopy in C. But make it so by "projecting" radially the line onto the circle. You can tell immediately that there is a problem with this construction when the line passes through the origin... that is to say, when f(x)=-x for some x! Make this idea precise by writing down the explicit formula for the homotopy described above and showing that the formula is a well defined homotopy iff f(x) is not -x *for some* x.

2. The cone CY is such that you can take any point on CY and get to any other point of CY by going up the cone and passing through the tip then going back down to the other point.

2. The cone CY is such that you can take any point on CY and get to any other point of CY by going up the cone and passing through the tip then going back down to the other point.

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- #3

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In both the two solutions you've presented I find the commen ground that you consider two maps are homotopic if every point they mapped can be connected by a curve.

But what I know is only the definition of homotopy:f is homotopic to g if there exists a map F:X*I-->Y such that F(x,0)=f(x) and F(x,1)=g(x) for all points x belongs to X.

How do we know they are equivalent?Thanks!

- #4

quasar987

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Of course to construct an homotopy from f to g it is not sufficient to just construct a bunch of path from f(x) to g(x) for all x.. these path must, "as a whole", move the space X around in a continuous manner. This is expressed in the requirement that the map F is continuous from the product space X x I to Y.

For instance, in 2., your homotopy, which it is convenient to consider as a bunch of path, will have to lift all the points of Y up to the tip of the cone more of less simultaneously, otherwise your homotopy will fail to be continuous.

As an illustration of what I mean, consider the case of a wannabe homotopy map F:X x I--> CY between f:X-->(Y x {0}) [itex]\subset[/itex] CY and g:X-->CY, such that as t goes from 0 to ¼, only one point of Y is lifted to the tip of the cone and all the others are maintained into place, and then as t goes from ¼ to ½, all the others are then lifted to the tip, and as t goes from ½ to 1, all the points run down to the appropriate value g(x). Well this is not a homotopy since it is not continuous: there is a "breaking" in the fabric of Y when that point if lifted up alone.

Note that I made some edits to post #2.

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Thank you,I understand it.:)

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being repd. by the antipode map.?.

I could find the analytic description of the homotopy:

H(x,t) = [tf+(1-t)g ]/|| tf+ (1-t)g||

(which breaks down at t=1/2, if f(x)=-g(x) )

where ||.|| is the standard Euclidean norm, but I am having trouble visualizing

what happens when a point x gets mapped to its antipode. I tried using a cylinder

to visualize it, ( the problem should become clear halfway through the cylinder,

corresponding to the fact that the problem happens at t=1/2) , but I still cannot see

it.

Any Ideas.?

- #7

quasar987

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This would mean that the fundamental group of the circle has two elements, which you know isn't true.Does this imply that there are only two homotopy classes of maps from the circle (S^n, actually, I think) ,to itself; one class being repd. by the identity, and the other class

being repd. by the antipode map.?.

I could find the analytic description of the homotopy:

H(x,t) = [tf+(1-t)g ]/|| tf+ (1-t)g||

(which breaks down at t=1/2, if f(x)=-g(x) )

where ||.|| is the standard Euclidean norm, but I am having trouble visualizing

what happens when a point x gets mapped to its antipode. I tried using a cylinder

to visualize it, ( the problem should become clear halfway through the cylinder,

corresponding to the fact that the problem happens at t=1/2) , but I still cannot see

it.

Any Ideas.?

Well, if we stick to the case g=identity for simplicity, and if x is a point such that f(x)=-x, then what does the "path" H(x,t) looks like? For 0<t<½, H(x,t)=-x, for t=½, it is undefined, and for ½<t<1, H(x,t)=x.

- #8

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Thanks for your patience. I am an analyst-in-exile; my strength is more on the

analysis and pointset areas , but I am trying to learn the algebraic and geometric

parts of topology.

Hope this is not too dumb.

I was working with the cyclinder again, and I wondered: if we took an extreme case

of the identity i(x)=x, and the antipode map a(x)=-x . Why couldn't we homotope

i(x) into a(x) by gradually rotating i(x) into a(x) , so that, at H(x,1) , we have

rotated i(x) by Pi.?. Then H(x,t) would be a gradual rotation, which is continuous.

What is wrong with this statement.?

- #9

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I understand, analytically, that my H(x,t) breaks down (meaning it is not continuous) at t=1/2 , for f(x)=i(x)=x, and g(x)= a(x)=-x. I was just trying to understand it more geometrically.

Besides, my H(x,t) is only one choice of homotopy, and there may be some

H'(x,t) that avoids the problem of H(x,t).

Thanks; if I don't get it this time, I will just go back to the drawing board myself

and figure out what I am missing.

- #10

quasar987

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Thanks for your patience. I am an analyst-in-exile; my strength is more on the

analysis and pointset areas , but I am trying to learn the algebraic and geometric

parts of topology.

Hope this is not too dumb.

I was working with the cyclinder again, and I wondered: if we took an extreme case

of the identity i(x)=x, and the antipode map a(x)=-x . Why couldn't we homotope

i(x) into a(x) by gradually rotating i(x) into a(x) , so that, at H(x,1) , we have

rotated i(x) by Pi.?. Then H(x,t) would be a gradual rotation, which is continuous.

What is wrong with this statement.?

Must there be something wrong?

The argument made above does not imply that the antipodal map is not homotopic to the identity. We simply said that if there is

Sure the particular homotopy we constructed to solve the problem breaks down for f=a the antipodal map, but that is not to say that we cannot find another one...which you did.

- #11

quasar987

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Sorry, one more comment:

I understand, analytically, that my H(x,t) breaks down (meaning it is not continuous) at t=1/2 , for f(x)=i(x)=x, and g(x)= a(x)=-x. I was just trying to understand it more geometrically.

Well, I don't know if this picture is what you would call more geometric, but recall what I said in post #2 as the intuitive picture for the homotopy:

"

Why can I tell immediately that there is a problem with this construction when the line passes through the origin? Because there is no natural way to project a diameter onto the circle. Do you project it to the half circle on the right or on the left of the diameter? In both cases, you obviously introduce a gross discontinuity.

- #12

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Just curious: if you have these examples at hand, could you please tell me

of two maps f:S^1-->S^1 that re not

homotopic to each other (except for the obvious cases of maps that wind

around the circle n and m times respectively.?. If i am not imposing too

much on you -- actually that ship sailed long ago, tho) .

If you don't have them offhand, don't worry, you have helped plenty.

- #13

quasar987

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As for this one, well the answer is that as is well known, the fundamental group of the circle is made up precisely of the homotopy classes of the maps z-->z^k for every integer k in

One could say that, up to homotopy, there are no non-homotopic maps S^1-->S^1 other than the z-->z^k for different values of k.

This said, an example could be: f=(a constant map) and g=(a map that, as u go around the circle counter-clockwise, g goes halfway around the circle, then backs up a quarter turn, then backs up again and complete the turn around the circle). Then f is homotopic to z-->z^0 and g is homotopic to z-->z^1, so f and g are an example of non-homotopic maps other than the canonical example of z-->z^k for different values of k.

Is this what you were asking for?

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