In 17:40..to prove \tildef^{-1}(V) open in T, you consider taking concatenation of paths from t to t^{\prime} with path t^{\prime} to t^{\prime\prime} and then you looked at the image of path joining t^{\prime} to t^{\prime\prime} under the map f. My question is instead of doing that , we can simply take the image of path joining t^{\prime} to t^{\prime\prime} under f because both t^{\prime} and t^{\prime\prime} lies in W and W is path connected by the construction. Does the question make sense? Or am I missing any subtle points?
I’m not sure I understand why we need to use the inclusion condition on the push-forwards to prove well-definedness of the lifting. Couldn’t we just lift the loop from X to Y? Wouldn’t the two halves of the lifted loop be themselves liftings of the two corresponding halves of the X loop (by definition of path lifting - along with path composition)? This would imply, in particular, that the inclusion condition is true of every path-connected, locally path-connected T, covering map p : Y -> X and continuous function f : T -> X. What am I missing?
Ok, I think I see why. Loops in T may be homotopically equivalent to the constant loop (this was not an issue in the case of path lifting or homotopy lifting, because the test spaces, [0,1] and [0,1]x[0,1] resp., were simply connected). As a result, it may not be possible to find a single elementary neighborhood which includes the pushforward of such a loop in its totality. Therefore, the lifting of such a loop to the covering space wouldn’t be a loop at all, making f tilde not well-defined. For instance, if T=S^1 and X is the torus, with f mapping T around the torus (either direction, it doesn’t matter). Then the lifting of (the pushforward of) a non-trivial loop in S_1 would be homotopically equivalent to a straight line in R^2. In particular, the starting and ending points would not be idetified, hence the ambiguity.
Hahaha! No worries, it’s not a stupid question, there’s so much to remember, it’s understandable. Here’s where you can find your answer: www.homepages.ucl.ac.uk/~ucahjde/tg/html/pi1-07.html
The nullhomotopy you mentioned is not end point homotopy is it? So this just says any map from S2 to T2 is freely homotopic to constant. Not end point homotopic.
It should be based homotopic. For your example, if you have a map from S^2 to T^2 sending the basepoint of S^2 to x and you pick a preimage y of x in the universal cover R^2 of T^2 then you get a map S^2 --> R^2 sending the basepoint to y. Now this map is based homotopic to the constant map at y, and this nullhomotopy (based at y) projects to a based nullhomotopy of your original map (because y projects to x).
Your videos really helped me study for my algebraic topology exam. Thank you so much
This video has been so helpful for me, thank you!!
In 17:40..to prove \tildef^{-1}(V) open in T, you consider taking concatenation of paths from t to t^{\prime} with path t^{\prime} to t^{\prime\prime} and then you looked at the image of path joining t^{\prime} to t^{\prime\prime} under the map f. My question is instead of doing that , we can simply take the image of path joining t^{\prime} to t^{\prime\prime} under f because both t^{\prime} and t^{\prime\prime} lies in W and W is path connected by the construction. Does the question make sense? Or am I missing any subtle points?
I think you’re right. In fact, in the notes, he does exactly what you said:
www.homepages.ucl.ac.uk/~ucahjde/tg/html/gal-01.html
I’m not sure I understand why we need to use the inclusion condition on the push-forwards to prove well-definedness of the lifting. Couldn’t we just lift the loop from X to Y? Wouldn’t the two halves of the lifted loop be themselves liftings of the two corresponding halves of the X loop (by definition of path lifting - along with path composition)?
This would imply, in particular, that the inclusion condition is true of every path-connected, locally path-connected T, covering map p : Y -> X and continuous function f : T -> X. What am I missing?
Ok, I think I see why. Loops in T may be homotopically equivalent to the constant loop (this was not an issue in the case of path lifting or homotopy lifting, because the test spaces, [0,1] and [0,1]x[0,1] resp., were simply connected). As a result, it may not be possible to find a single elementary neighborhood which includes the pushforward of such a loop in its totality. Therefore, the lifting of such a loop to the covering space wouldn’t be a loop at all, making f tilde not well-defined.
For instance, if T=S^1 and X is the torus, with f mapping T around the torus (either direction, it doesn’t matter). Then the lifting of (the pushforward of) a non-trivial loop in S_1 would be homotopically equivalent to a straight line in R^2. In particular, the starting and ending points would not be idetified, hence the ambiguity.
Why do elementary sheets form a base for the topology on Y?
I hope I get an answer to this incredibly stupid question but anyway. What do we mean by p*
I think that is the Homomorphisms Induced by the covering map.
Take a look at www.maths.ed.ac.uk/~v1ranick/papers/viro.pdf section 36
Hahaha! No worries, it’s not a stupid question, there’s so much to remember, it’s understandable.
Here’s where you can find your answer:
www.homepages.ucl.ac.uk/~ucahjde/tg/html/pi1-07.html
The nullhomotopy you mentioned is not end point homotopy is it?
So this just says any map from S2 to T2 is freely homotopic to constant. Not end point homotopic.
It should be based homotopic. For your example, if you have a map from S^2 to T^2 sending the basepoint of S^2 to x and you pick a preimage y of x in the universal cover R^2 of T^2 then you get a map S^2 --> R^2 sending the basepoint to y. Now this map is based homotopic to the constant map at y, and this nullhomotopy (based at y) projects to a based nullhomotopy of your original map (because y projects to x).
very nice explanation
Glad you found it useful!