- #1
Bacle
- 662
- 1
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.
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.