Does path connected not imply locally path connected? In the general lifting criterion, both of these things are assumed, but I thought path connected implies locally path connected.
Agreed. I'm confused when he was doing the final part of the proof trying to show that f tilda is continuous. He picked a sheet V over U, which is fine, but then he suddenly started to pretend that V was inside Y??? Like what? And this just went on
@@berserker8884 in the book \tilde U is used for \tilde X and V for Y. Apparently, the lecturer here got confused and used V for \tilde X as he did before, but then also went for V in Y like in the book
Yes, but still not enough, since you want to use locally path-connectedness. In fact, y' should live in an open subset of f^{-1}(U) that is path connected.
What I think is the the ‘a’ which is on the bottom left of the top drawing (second from the top on left) should have its arrow point down instead of up. Then each of the three base points in the top drawing will have an ‘a’ going into it and an ‘a’ going out of it. That being said, this space S^1 v S^1 is just a wedge sum of two circles; it doesn’t require arrows. The whole thing can be drawn with no arrows.
im thinking the theorem indicates a bijection from coset to p^-1(x0) would not work for the left cosets since that subgroup isnt normal
At 1:01:42 g' dot g bar, not g' dot g
Does path connected not imply locally path connected? In the general lifting criterion, both of these things are assumed, but I thought path connected implies locally path connected.
no it is not true , consider comb space , it is path connected but not locally path connected
The conjunction of connected and locally path connected does imply path connected.
the proof of 'general lifting criterion' seems really strange.. especially the notation is abused.
Agreed. I'm confused when he was doing the final part of the proof trying to show that f tilda is continuous. He picked a sheet V over U, which is fine, but then he suddenly started to pretend that V was inside Y??? Like what? And this just went on
@@berserker8884 in the book \tilde U is used for \tilde X and V for Y. Apparently, the lecturer here got confused and used V for \tilde X as he did before, but then also went for V in Y like in the book
1:08:55 This should say “y’ in f^-1(U)”, where this is an open subset of Y.
Yes, but still not enough, since you want to use locally path-connectedness. In fact, y' should live in an open subset of f^{-1}(U) that is path connected.
yes,starting from where you say, and substitute some of the V (not all) to f preinmage of U, statement will be right.
Ew why is that chalkboard brown?
8:30 gaaahhhhhh the left top a should have opposite orientation
very confusing, yes
i think the the left bottom 'a' should have opposite orientation.... or else both the top 'a' should change their orientations
What I think is the the ‘a’ which is on the bottom left of the top drawing (second from the top on left) should have its arrow point down instead of up. Then each of the three base points in the top drawing will have an ‘a’ going into it and an ‘a’ going out of it.
That being said, this space S^1 v S^1 is just a wedge sum of two circles; it doesn’t require arrows. The whole thing can be drawn with no arrows.