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I was reading that the relation on $S^{n-1} \times I$ that identifies $(x,1) \sim (y,1)$ for all $x,y \in S^{n-1}$ and also $(x,0) \sim (y,0)$ for all $x,y \in S^{n-1}$ give us the suspension of $S^{n-1}$ and I want to explain for myself what will happen in the universal property of the quotient topology if we carried this quotient in two steps ( one for $(x,1) \sim (y,1)$ and then one for $(x,0) \sim (y,0)$) not just one as in the definition. I am a little bit rusty about the universal property of the quotient topology. I can see a version of it here Proof of the universal property of the quotient topology. and here universal property in quotient topology and I am revising it now but still I can not see how will apply it one time to two equivalence relations at the same time while another time to only one equivalence relation.

Any help will be greatly appreciated!

Hope
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1 Answers1

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Strictly speaking you'd need to change notation slightly, e.g. if you quotient by $(x,1)\sim(y,1)$ first then I might pedantically want to see $[(x,0)]\sim[(y,0)]$ in the second, to distinguish elements of $S^{n-1}\times I/\text{first relation}$ from elements of $S^{n-1}\times I$.

Here is the only technicality that you need:

  • The composition of two quotient maps $q_1,q_2$ is again a quotient map

Given this, we can figure what happens when we do things one by one. Note in general it's not clear how the two relations interact, and it may not be possible to easily split them into "step 1 and step 2"; it's very important that the first relation is somehow 'disjoint' from the second relation.

So we get a quotient map $S^{n-1}\times I\to S^{n-1}\times I/\sim_1\to(S^{n-1}\times I/\sim_1)/\sim_2$, and we want to realise what the resulting relation is. If it's the correct one, the suspension relation, then that exactly says that doing it "one by one" gives us a space (homeomorphic to) the suspension.

Ok, fix $(x,t)$ and $(y,t')$ in $S^{n-1}\times I$. They are related in the composition iff. $[(x,t)]\sim_2[(y,t')]$ i.e. iff. either $[(x,t)]=[(x',0)],[(y,t)]=[(y',0)]$ for some $x',y'$ or $[(x,t)]=[(y,t')]$. By inspection of how $\sim_1$ works, we realise the first case occurs iff. $t=0=t'$ (and $x=x',y=y'$) i.e. iff. our original $(x,t),(y,t')$ both lived in $S^{n-1}\times\{0\}$. We also realise the second case happens iff. $(x,t),(y,t')$ both live in $S^{n-1}\times\{1\}$.

Therefore, the composite quotient map does exactly what we want it do: it only identities points in $S^{n-1}\times\{0\}$ and $S^{n-1}\times\{1\}$ and it identifies the top and bottom spheres in distinct ways. (we want to avoid $(x,0)\sim(y,1)$, for instance.

FShrike
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  • I can not see how you are applying the universal property that I gave in the first link for the first relation and after that for the second relation. Can you show the concrete steps please? – Hope Jan 27 '24 at 23:18
  • @Hope I haven't used the universal property once, actually. I've only used the explicit construction of the quotient spaces. Ok, that's not exactly true. I use the special case of the universal property exactly once to say: if there are quotient maps $X\to Y,X\to Z$ inducing the same relation on $X$, then $Y\cong Z$. But that's the least important/interesting part of my answer, all the relevant details lie in the other text. – FShrike Jan 27 '24 at 23:19
  • it seems like you are saying that the solution is not by applying the universal property two times right? – Hope Jan 27 '24 at 23:23
  • How about if we used what you used in your answer for the two equivalence relations at the same time? I wanna get an answer from this and another answer from applying the universal property twice and then see that they are equal. That will relief my brain more. Can you help me in that? – Hope Jan 27 '24 at 23:24
  • @Hope I encourage you to be more specific. Is there something in particular in my answer that you don't understand? Morally, any solution to this will use the fact quotient spaces are determined by their relations, and the "composition" of the relations for each step - in this case! - corresponds to the overall suspension relation you want. Checking the universal property actually applies comes down to doing exactly this, checking what the induced relations are, so I'm not sure what more you want – FShrike Jan 27 '24 at 23:38
  • let me read your solution again and I will let you know about my questions – Hope Jan 28 '24 at 01:14
  • at the end of the first paragraph .... Did you meant to complete your statement "to distinguish elements of $S^{n-1}\times I/\text{first relation}$ from elements of $S^{n-1}\times I$." By the following " from elements of $S^{n-1}\times I/\text{Second relation}$" – Hope Jan 28 '24 at 01:47
  • I did not get what exactly are you doing in this paragraph "Ok, fix $(x,t)$ and $(y,t')$ in $S^{n-1}\times I$. They are related in the composition iff. $[(x,t)]\sim_2[(y,t')]$ i.e. iff. either $[(x,t)]=[(x',0)],[(y,t)]=[(y',0)]$ for some $x',y'$ or $[(x,t)]=[(y,t')]$. By inspection of how $\sim_1$ works, we realise the first case occurs iff. $x=x',y=y'$ i.e. iff. our original $(x,t),(y,t')$ both lived in $S^{n-1}\times{0}$. We also realise the second case happens iff. $(x,t),(y,t')$ both live in $S^{n-1}\times{1}$." Could you elaborate please? – Hope Jan 28 '24 at 01:58
  • what "inspection" are you referring to in the paragraph I mentioned in my previous comment? – Hope Jan 28 '24 at 01:59
  • @Hope (To your first comment) No, I meant what I said. When we are defining the second relation, we need to discuss elements of $S^{n-1}\times I/\text{first relation}$. To do so carefully, we should bear in mind that elements of this quotient space are not the same thing as elements of $S^{n-1}\times I$, so strictly speaking we should use different notation. – FShrike Jan 28 '24 at 02:09
  • @Hope To your second comment: there are several statements there, it would help if you mentioned which sentences don't make sense (and why). I'm essentially just following the definitions of the relations! To your third comment: the "inspection" is just... thinking. There's not really a substitute. However, I made an omission: I should have emphasised "$t=0=t'$" as the main point, rather than "$x=x',y=y'$". I'll edit the post and hopefully it'll make more sense then – FShrike Jan 28 '24 at 02:12
  • Maybe the wall of notation has confused you. The idea is just this: yes, it's ok to do it step by step. Why is it ok? Essentially because, if you do it step by step, you only ever apply the $\sim_1$ and $\sim_2$ relations, there is no "cross talk", nothing gets messed up, ... the intuition for what the resulting relation should be is correct; it is indeed the suspension relation. And you just check it! You need to check that $[a]\sim_2[b]$ in the $\sim_1$-quotient iff. $a\sim b$ according to the suspension relation. That's saying: "the step by step relation equals the combined relation". – FShrike Jan 28 '24 at 02:19
  • in the line of the word either, did you meant $[(y,t')] = [y',0]$? – Hope Jan 31 '24 at 10:46
  • in the either or sentence are you looking at changing x and y and then fixing $t=0$ and the other case fixing x and y and changing t? How did you get these two cases? – Hope Jan 31 '24 at 10:50
  • @Hope I meant what I wrote, $[(y,t’)]=[(y’,0)]$. I’m not changing or fixing anything. I’m following the definition of the equivalence relations. I’m given $[(x,t)]\sim_2[(y,t’)]$ and now I have to figure out what exactly that means. I’m just using the definition of $\sim_2$ (identify all points in the image of $S^1\times{0}$). – FShrike Jan 31 '24 at 12:56