I'm a stupid layperson who has trouble understanding any other video (or book or anything) on topology and your videos are the only ones that help me understand.
THANK YOU VERY MUCH!! Not just that I feel like I understand quotent spaces now, but equivalence relation has always been so abstract to me, I kinda got it but not really and now all of it seems so clear to me. You just made my day :)
At 4:58 you say the preimage of Y is X. Many books say this is true because q is surjective. Do we need q to be surjective? I don't understand why we need q to be surjective to conlude that the preimage of Y is X. If that is true, then where did we use surjectivity at all?
You are correct. The way I wrote it in the video, we would identify all points (x,y) with (x,-y), which would indeed yield a half disk, which is homeomorphic to the original disk. I meant for the equivalence relation only to apply to boundary points, but forgot to write this explicitly. Good catch!
In the example demonstrating how to glue the endpoints of the unit interval, J = [0, 1], to make its quotient space homeomorphic to S^1, I would like to simply note that the map x \mapsto (cos(2pi x), sin(2pi x)) from J to S^1 is not an open map because the interval [0, a) (a
That seems like an interesting way to view the problem of that example. I'm not sure how well this type of reasoning would generalize to more complicated examples where there is more than one problem point. For me it is easier to think of walking a path in the two spaces simultaneously and thinking about what parts have to be identified. Formally, if you have any continuous path on S^1, then a homeomorphism to J/~ can be used to transport the path to a continuous path on J/~. If you walk around the circle you see that you need to identify 0 and 1 in J, otherwise the path would jump. This is one way to see which parts have to be identified, but I don't think it is sufficient.
Can you talk about how to show homeomorphism between a quotient map of a space to a target space also, when showing such homeomorphism, do you always need to construct a specific function using parametric equations, example of such case is defining an equivalence relation by identifying the endpoints of an interval and that is homeomorphic to a circle. But the case of homeomorphisms for identification of a subspace and the Klein bottle or mobius band are presented not in such explicit manner. Thanks
Yes, I will discuss this type of thing more in parts 2 and 3, which I will record once I can find the time. If those future videos don't answer your questions, feel free to ask again in the comments there.
Why is the second from last example a cone. We haven't acted on say (x, 0.5) so that 0. 5 should be all of X still. All the cross sectional cuts except the top are similar in the picture but we only collapsed (x, 0) layer. Please explain
Yes, it is maybe a bit confusing to call it a cone. What I mean that it is topologically a cone, i.e. homeomorphic to one. You can homeomorphically shrink down the cross sections of a cylinder to whatever width you want. What you can't do, however, is homeomorphically shrink the bottom circular cross section to a point. Thus we need to make the identification in the quotient. But calling it a cone is perhaps misleading, because many shapes that are not conic are homeomorphic to cones. For example, a disc is also homeomorphic to a cone (by squashing it), as is a bowl shape.
In the wedge sum example, is it correct to say that to have the wedge sum of spaces X and Y with the base point choice x in X and y in Y, we take the equivalence closure of the relation R = {((x, 1), (y, 2))}, i.e., the smallest eqv. relation containing R, to form a quotient space? I am asking about this because Wikipedia en.wikipedia.org/wiki/Wedge_sum just chose R = {(x, y)}, which I think is not quite technically correct because it is not a relation on the disjoint union though it is at least semantically acceptable.
Yes, you are right. One way to realize the disjoint union is to "tag" each set with an index. So e.g X becomes X x {1} and Y becomes Y x {2}. In this case we would want to identify (x,1) with (y,2). Because we can always add a tag, one can also just assume without loss of generality that X and Y are disjoint to begin with. But you are right that one needs to watch out when making these types of implicit identifications and it is good that you are thinking this through precisely. In concrete cases one does actually need to distinguish things.
@@ahalyageorgeclinton486 The part you are confused about is an argument by contradiction which shows that if we included any more open sets in the quotient topology, the function q would no longer be continuous. Call the quotient topology Q. Take a subset S not in Q. Then q-1(S) cannot be open, since otherwise S would be in Q by definition (Q contains all subsets U of Y such that q-1(U) is open in X). Hence there would be an open set in this larger topology with a pre-image that is not open, which means q would not be continuous. Hence Q is the largest topology we can put on Y which makes q continuous. To answer your second question: It is not the case that any continuous surjective map is a quotient map if you fix the topologies on X and Y in advance. I talk about this more in Quotient Spaces II. For example, let X := {a,b} with the discrete topology (every subset is open) and Y := {c,d} with the trivial topology (only Y and emptyset are open). Let q map a -> c and b -> d. Then q is continuous and surjective, but q-1({c}) is open in X even though {c} is not open in Y. Hence the topology on Y does not satisfy the definition of the quotient topology. Of course if you are given any surjective continuous map q: X -> Y, you can always put a new topology on Y which is the quotient topology (because q is in particular a surjective function). But this need not have anything to do with the old topology that was on Y.
I'm a stupid layperson who has trouble understanding any other video (or book or anything) on topology and your videos are the only ones that help me understand.
Thank you so much for these videos, this is a hard topic to teach and you made it so great!
You are very welcome!
Thank you for this exhaustive presentation.
Amazing video, thank you very much you teach very good.
THANK YOU VERY MUCH!! Not just that I feel like I understand quotent spaces now, but equivalence relation has always been so abstract to me, I kinda got it but not really and now all of it seems so clear to me. You just made my day :)
Great to hear that has clicked for you!
At 4:58 you say the preimage of Y is X. Many books say this is true because q is surjective. Do we need q to be surjective? I don't understand why we need q to be surjective to conlude that the preimage of Y is X. If that is true, then where did we use surjectivity at all?
Excellent lectures mauris. Will see this series completely and recommend to my students as well. Thank you
Thank you so much. Really enjoying these lectures.
You are very welcome.
You are a saviour !
Greatly enlightening! Thanks for sharing
This guy is a genius.
Awesome ❤
Hi, I have a question about the second example. Can it be the half disk since the equivalent relation you defined is (x,y)~(x,-y).
You are correct. The way I wrote it in the video, we would identify all points (x,y) with (x,-y), which would indeed yield a half disk, which is homeomorphic to the original disk. I meant for the equivalence relation only to apply to boundary points, but forgot to write this explicitly. Good catch!
In the example demonstrating how to glue the endpoints of the unit interval, J = [0, 1], to make its quotient space homeomorphic to S^1, I would like to simply note that the map x \mapsto (cos(2pi x), sin(2pi x)) from J to S^1 is not an open map because the interval [0, a) (a
That seems like an interesting way to view the problem of that example. I'm not sure how well this type of reasoning would generalize to more complicated examples where there is more than one problem point. For me it is easier to think of walking a path in the two spaces simultaneously and thinking about what parts have to be identified. Formally, if you have any continuous path on S^1, then a homeomorphism to J/~ can be used to transport the path to a continuous path on J/~. If you walk around the circle you see that you need to identify 0 and 1 in J, otherwise the path would jump. This is one way to see which parts have to be identified, but I don't think it is sufficient.
Can you talk about how to show homeomorphism between a quotient map of a space to a target space also, when showing such homeomorphism, do you always need to construct a specific function using parametric equations, example of such case is defining an equivalence relation by identifying the endpoints of an interval and that is homeomorphic to a circle. But the case of homeomorphisms for identification of a subspace and the Klein bottle or mobius band are presented not in such explicit manner. Thanks
Yes, I will discuss this type of thing more in parts 2 and 3, which I will record once I can find the time. If those future videos don't answer your questions, feel free to ask again in the comments there.
the videos are so great.are you going to update some vedios about the compactness and connectivity？i am looking forward for it
I am very happy you are enjoying the videos! I will start discussing compactness and connectivity two videos from now, so in about 1 month.
@@mariusfurter The videos are amazing! Compactness would be a very welcome topic.
Why is the second from last example a cone. We haven't acted on say (x, 0.5) so that 0. 5 should be all of X still. All the cross sectional cuts except the top are similar in the picture but we only collapsed (x, 0) layer. Please explain
Yes, it is maybe a bit confusing to call it a cone. What I mean that it is topologically a cone, i.e. homeomorphic to one. You can homeomorphically shrink down the cross sections of a cylinder to whatever width you want. What you can't do, however, is homeomorphically shrink the bottom circular cross section to a point. Thus we need to make the identification in the quotient.
But calling it a cone is perhaps misleading, because many shapes that are not conic are homeomorphic to cones. For example, a disc is also homeomorphic to a cone (by squashing it), as is a bowl shape.
Thank you for explaining!
Can you make playlist on algebraic topology ???!! pleeeease
It's possible that I might make some videos on homotopy and the fundamental group in the next months but I can't make any promises.
In the wedge sum example, is it correct to say that to have the wedge sum of spaces X and Y with the base point choice x in X and y in Y, we take the equivalence closure of the relation R = {((x, 1), (y, 2))}, i.e., the smallest eqv. relation containing R, to form a quotient space? I am asking about this because Wikipedia en.wikipedia.org/wiki/Wedge_sum just chose R = {(x, y)}, which I think is not quite technically correct because it is not a relation on the disjoint union though it is at least semantically acceptable.
Yes, you are right. One way to realize the disjoint union is to "tag" each set with an index. So e.g X becomes X x {1} and Y becomes Y x {2}. In this case we would want to identify (x,1) with (y,2). Because we can always add a tag, one can also just assume without loss of generality that X and Y are disjoint to begin with. But you are right that one needs to watch out when making these types of implicit identifications and it is good that you are thinking this through precisely. In concrete cases one does actually need to distinguish things.
@@mariusfurter Thank you for the confirmation :) I think I need to make myself accustomed to such simplified expressions.
Thanks for this 😀
You're very welcome :)
Great video, are you going to continue?
Thanks! Yes, as the numbering suggests I will continue, but I've been very busy lately. I hope to record and post the next video within the next week.
@@mariusfurter Please do so soon as am desperately waiting for the next video
Sorry for the long wait. I have finally found the time to record the next video! I should have it edited and uploaded by tomorrow.
thanks a lot
Why s is q-1(s) is not open in X
Could you post the time-code to where you are referring to?
3.02 time your video
And I am week in topology so, I will ask you. If q is continues with Surjective, is it quotient.
@@ahalyageorgeclinton486 The part you are confused about is an argument by contradiction which shows that if we included any more open sets in the quotient topology, the function q would no longer be continuous. Call the quotient topology Q. Take a subset S not in Q. Then q-1(S) cannot be open, since otherwise S would be in Q by definition (Q contains all subsets U of Y such that q-1(U) is open in X). Hence there would be an open set in this larger topology with a pre-image that is not open, which means q would not be continuous. Hence Q is the largest topology we can put on Y which makes q continuous.
To answer your second question: It is not the case that any continuous surjective map is a quotient map if you fix the topologies on X and Y in advance. I talk about this more in Quotient Spaces II. For example, let X := {a,b} with the discrete topology (every subset is open) and Y := {c,d} with the trivial topology (only Y and emptyset are open). Let q map a -> c and b -> d. Then q is continuous and surjective, but q-1({c}) is open in X even though {c} is not open in Y. Hence the topology on Y does not satisfy the definition of the quotient topology.
Of course if you are given any surjective continuous map q: X -> Y, you can always put a new topology on Y which is the quotient topology (because q is in particular a surjective function). But this need not have anything to do with the old topology that was on Y.
i'm officially worshipping you