I'm a bit confused by the definition of a good pair. Is there a closed subspace A of X (topological space) that is not a deformation retract of a neighborhood in X? The definition suggests that the answer is yes but I don't see how this is possible.
On a second thought, the pair of unit interval and the Cantor set seems to have such a property - the Cantor set being closed but not a deformation retract since any neighborhood of any point in the Cantor set contains infinite number of points in it. Is this observation true?
There are much easier examples. Take X to be the unit interval and take A to be two distinct points. Then A is closed, but cannot be a deformation retract of a neighborhood in X, since any neighborhood in X is connected, while A is not. (A deformation retract induces a homotopy equivalence, which preserves, e.g., the number of connected components.) P.S. A "good pair" is an example of the much more general concept of a "cofibration." Reading about such things will show you where the idea of a good pair originates from.
Where can I find the proof that the prism decomposition that was used is indeed both a cover, and one where the interiors don't overlap? I've searched the term "prism" in both Hatcher and J. P. May and I haven't found anything.
@@rancidrufus There's no proof of what I asked there. It states the fact in passing, but does not prove it, nor use it. I'm asking for a proof of that specific geometric fact.
@@thephysicistcuber175 Are you referring to a proof that the space Delta^n x I is indeed the union of the simplices? The proof is, barebones, something like this: Take some point in Delta^n x I. Consider its barycentric coordinates relative to each of the constructed simplices. If you run through the math, you'll find that at least one of the coordinates has to have all positive values, which in turn means it must sit inside one of the simplices.
A good pair of lectures.
39:45 exact sequences
I'm a bit confused by the definition of a good pair. Is there a closed subspace A of X (topological space) that is not a deformation retract of a neighborhood in X? The definition suggests that the answer is yes but I don't see how this is possible.
On a second thought, the pair of unit interval and the Cantor set seems to have such a property - the Cantor set being closed but not a deformation retract since any neighborhood of any point in the Cantor set contains infinite number of points in it. Is this observation true?
There are much easier examples. Take X to be the unit interval and take A to be two distinct points. Then A is closed, but cannot be a deformation retract of a neighborhood in X, since any neighborhood in X is connected, while A is not. (A deformation retract induces a homotopy equivalence, which preserves, e.g., the number of connected components.) P.S. A "good pair" is an example of the much more general concept of a "cofibration." Reading about such things will show you where the idea of a good pair originates from.
Where can I find the proof that the prism decomposition that was used is indeed both a cover, and one where the interiors don't overlap? I've searched the term "prism" in both Hatcher and J. P. May and I haven't found anything.
Its in Theorem 2.10 on page 111,112.
@@rancidrufus There's no proof of what I asked there. It states the fact in passing, but does not prove it, nor use it. I'm asking for a proof of that specific geometric fact.
@@thephysicistcuber175 Are you referring to a proof that the space Delta^n x I is indeed the union of the simplices? The proof is, barebones, something like this: Take some point in Delta^n x I. Consider its barycentric coordinates relative to each of the constructed simplices. If you run through the math, you'll find that at least one of the coordinates has to have all positive values, which in turn means it must sit inside one of the simplices.
@@ExplosiveBrohoof Ok, and do you do something similar to show that the interiors don't overlap?
@@thephysicistcuber175 The interiors don't need to be disjoint, I don't think, and I believe in general they won't be.
Homology is dual to co-homology.
Homotopic equivalence = duality!