4.02 The homotopy extension property
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- čas přidán 12. 09. 2018
- We introduce the homotopy extension property (HEP). We show that if a contractible subspace has the HEP then we can crush it to a point without changing the homotopy type of the ambient space.
Always crystal clear ❤
i watched many of your videos. It helps a lot. I must say that you are an amazing teacher.
Glad you found them useful!
Thanks again for another great video! I’m wondering if the definition of HEP could be simplified by letting X=Y, F=id_X and A x [0,1] -h-> A (where Y on the rhs became A). It seems to be what you’re actually using in this particular proof and it would require one quantifier less (for all F …).
I’m not sure if it would be equivalent to your definition in general though. In other words, is every homotopy of type A x [0,1] -h-> B always decomposable into a homotopy of type A x [0,1] -h-> A, followed by some A -f-> B ?
Ok, it wasn’t too hard to check that the answer to my question is "no, they’re not equivalent”. It suffices to let A be the singleton set and B=[0,1] , with h isomorphic to id_B. Then A x [0,1] -h'-> A can only be the constant map, and there is no A -f-> B such that h = f\circ h’.
Then the true definition of HEP (the one you gave) is stronger than my “simplified variant”. But I’m still wondering if there are any interesting cases that mine is too weak to capture.
Thank you so much
The example of the pairs what you have mentioned in the beginning of this video are actually CW pairs. Right? For the first example you are attaching two one cells to A and A being a line is contractible. For the second case you are attaching a three cell to A. So in both the cases X is a CW complex which contains A as a subcomplex and hence in both the cases (X,A) are CW pairs. Since CW pairs always satisfy homotopy extension property we are done! Does it make sense what I have written?
You can certainly give these guys cell structures so that A is a subcomplex. For example, for the theta graph you can think of theta as a graph with two vertices (0-cells) and 3 edges (1-cells) and then A is a subcomplex consisting of two 0-cells and one 1-cell. In the second example you can take 0-cells at the north and south poles, a 1-cell down the Greenwich meridian, a 2-cell covering the rest of the Earth, and then the additional 1-cell that goes up into space connecting the north and south poles; again, A is one 1-cell and two 0-cells. Indeed it's true that CW pairs have the HEP, but at this point in the course I hadn't proved that fact (that's the next video).
I think there is a confusion with the definition of 'contractible space'. If I'm not mistaken (I might be, I'm just learning) what has been proved here is that if A deformation retracts to a point and (X,A) has the HEP then X is homotopically equivalent to X/A. But A could be contractible without deformation retracting to a point, in which case this proof doesn't seem to work, yet the proposition still holds (but the proof should be modified).
Thanks for your comment. I hadn't actually ever thought about the distinction you're making, but the proof doesn't require a deformation retraction to a point. I think the point you're making is that the homotopy from the identity to the constant map might move the basepoint for a contractible space (I believe this is the only distinction between contractibility and deformation retraction to a point). This doesn't enter into the proof above.
@@jonathanevans27 yes, you're totally right, the confusion was mine. Thanks for making it clear!
Amazing video. It helped me a lot in the uni, seriously. I just don't get to understand how the existance of that function g we were looking for is deduced from the fact H_1(A) = {a}. Of course, as H_t is an homotopy from X to X s.t. H_0 = Id_X, we would love to be able to write H_1(A) = g o q for some g:X/A-->X, but how do we get the existance of g? Just by inversing the relation to write g explicitly as g = H_1 o p^-1 and verify that this g is correct?
H_1 sends A to a point, so factors through the quotient map X-->X/A. This is a key property of the quotient topology, see the earlier video czcams.com/video/4aI38YLSxFw/video.html
I'm heading there, thanks.
Basic question, but how exactly do we know \bar{H_t} is a homotopy?
I guess I was being sloppy writing my homotopies with t as a subscript instead of an argument (I find this more intuitive). You could argue as follows: there's a homotopy H: X \times [0,1] --> X and postcomposing with q gives you a map q\circ H: X \times [0,1] --> X/A. This map is constant on A \times [0,1] (basically as explained in the video) so factors as \bar{H}\circ (q,id) for some continuous map \bar{H}: X/A \times [0,1] --> X/A. Here I'm quotienting by the equivalence relation (a,t) ~ (a',t) a,a' \in A instead of the coarser relation (a,t) ~ (a',t') for a,a' \in A, t,t' \in [0,1]. Now \bar{H} is your homotopy (and \bar{H}(x,t) = \bar{H}_t(x)).