We show that a closed subcomplex of a CW complex has the HEP. From this, we deduce that any 1-dimensional CW complex is homotopy equivalent to a wedge of circles.
Thanks a lot for your video, it is very helpful ! I have a question however. At 11:00, you use the lemma to give a proof of the theorem ... by induction. But it is a problem, isn't it ? Because the induction can be used if we had a countable number of cells. However, it is not the case sometimes ...
@@jonathanevans27 Well, I'm not quite familiar with CW complexes, but all definitions I saw (especially the definition from Hatcher) take a family of morphisme phi_alpha to construct it, and it is never precised if the family is countable or not :x Thanks for your answer anyway !
A CW complex X is a union of cells, which are the images of maps from the k-dimensional disc into X. An open cell is where you restrict this map to the open disc (i.e. {x : |x|
One thing im confused about is, in the proof of the first statement, did you use the fact that A is closed? perhaps you need it to explicitly construct the map later on using pasting lemma.
Great question. The fact that you're attaching closed cells is implicit throughout the whole argument (e.g. when I draw the red part of the boundary of e x [0,1]), but I'd have to think quite hard to figure out exactly what role that's playing in the proof. However, I think it is crucial: you can't contract open cells without changing the homotopy type (which you could if they satisfied the HEP). Think about the 2-sphere with one 0-cell and one 2-cell: if you contract the contractible open 2-cell then you just end up with a 2-point space {a,b} with the non Hausdorff topology {\emptyset, {a}, {a,b}}.
Isn't subcomplex of a CW complex always closed? In Hatcher's book I find the argument that a subcomplex of a CW complex is closed and that can be proved inductively by showing that intersection of the subcomplex with the skeletons are closed. May I know which book did you follow while preparing these videos? Thanks.
I think you're right, I just didn't want people going away thinking you could contract an open 1-cell. For example, if you take the circle (with one 0-cell and one 1-cell) and collapse the open 1-cell then you end up with a 2 point space {0,1} with the topology where \emptyset, {0} and {0,1} are open.
Great video, just one thing I don't understand: at 10:00 it doesn't seem to me that all points in e×{0} U de×[0,1] are either in the domain of h or in the domain of F, for instance if you take any point in de (border of e) which is not in A and you lift it, say by 1/2, it is not in the domain of h nor in the domain of F. I would preciate any clarification on the subject
Good question: it's generally tricky to explain these kinds of construction properly without the notation getting very messy. In this case, you're attaching e to A, so all points in the boundary of e map to A. This means the points you're worried about don't exist, so all is well with the world.
![The moment we stopped understanding AI [AlexNet]](/img/n.gif)
The last fact is a reformulation of the fact that a graph has a maximal tree and trees are contractible. I am in love with your videos .
Thanks for explaining this so clearly, this was really helpful
Thanks a lot for your video, it is very helpful !
I have a question however. At 11:00, you use the lemma to give a proof of the theorem ... by induction. But it is a problem, isn't it ? Because the induction can be used if we had a countable number of cells. However, it is not the case sometimes ...
I never thought about CW complexes with uncountable numbers of cells, but I'm sure you can figure out a way to extend the proof if you need it.
@@jonathanevans27 Well, I'm not quite familiar with CW complexes, but all definitions I saw (especially the definition from Hatcher) take a family of morphisme phi_alpha to construct it, and it is never precised if the family is countable or not :x
Thanks for your answer anyway !
Thanks for the video! Could you please elaborate on the closed cells？Are they closed in the sense of topology or in the sense of manifold?
A CW complex X is a union of cells, which are the images of maps from the k-dimensional disc into X. An open cell is where you restrict this map to the open disc (i.e. {x : |x|
One thing im confused about is, in the proof of the first statement, did you use the fact that A is closed? perhaps you need it to explicitly construct the map later on using pasting lemma.
Great question. The fact that you're attaching closed cells is implicit throughout the whole argument (e.g. when I draw the red part of the boundary of e x [0,1]), but I'd have to think quite hard to figure out exactly what role that's playing in the proof. However, I think it is crucial: you can't contract open cells without changing the homotopy type (which you could if they satisfied the HEP). Think about the 2-sphere with one 0-cell and one 2-cell: if you contract the contractible open 2-cell then you just end up with a 2-point space {a,b} with the non Hausdorff topology {\emptyset, {a}, {a,b}}.
Isn't subcomplex of a CW complex always closed? In Hatcher's book I find the argument that a subcomplex of a CW complex is closed and that can be proved inductively by showing that intersection of the subcomplex with the skeletons are closed.
May I know which book did you follow while preparing these videos? Thanks.
I think you're right, I just didn't want people going away thinking you could contract an open 1-cell. For example, if you take the circle (with one 0-cell and one 1-cell) and collapse the open 1-cell then you end up with a 2 point space {0,1} with the topology where \emptyset, {0} and {0,1} are open.
The Sierpinski space :)
Great video, just one thing I don't understand: at 10:00 it doesn't seem to me that all points in e×{0} U de×[0,1] are either in the domain of h or in the domain of F, for instance if you take any point in de (border of e) which is not in A and you lift it, say by 1/2, it is not in the domain of h nor in the domain of F. I would preciate any clarification on the subject
Good question: it's generally tricky to explain these kinds of construction properly without the notation getting very messy. In this case, you're attaching e to A, so all points in the boundary of e map to A. This means the points you're worried about don't exist, so all is well with the world.
Ahhh I think I see it now, The drawing got me confused. Thank you very much