what theorem are you using to conclude that if it is sure tive then it is a quotient map?

what do you mean by fibers? Do you mean inverse images?

@Intuition A fibre is common terminology for some preimage of a point, $f^{-1}\{p\}$. Recall fibres partition the space and quotient maps correspond to the relation induced by their fibres, $a\sim b$ iff they live in the same fibre. Now let $f:X\to Y$ be a surjective map with $X$ compact, $Y$ Hausdorff. Let $U$ be an $f$-saturated open in $X$. $K:=X\setminus U$ is closed hence compact; $f(K)$ is compact hence closed. By surjectivity and assumption on $U$, $f(U)=Y\setminus f(K)$ is then open; $f$ is a quotient map.

what do you mean by saturated?

@Intuition A set $U$ is saturated w.r.t an equivalence relation $\sim$ iff. it has the following property: $u\in U$ and $v\sim u$ implies $v\in U$. That is, $U$ contains all equivalence classes of its elements, $U=\bigcup_{u\in U}[u]_\sim$. It is well known a continuous function $f$ is a quotient map (in the topological sense) iff. it is surjective and maps $f$-saturated open sets to open sets

are you using this theorem in the idea of your solution above: "If a continuous bijection has as its domain a compact space and its codomain is Hausdorff, then it is a homeomorphism."

@Intuition No. But it's an extremely similar theorem with an almost identical proof. In fact you could view the homeomorphism theorem as a special case of the theorem I gave; in the homeomorphism case, we have a continuous bijection, the theorem I stated shows it's a quotient map, and injective quotient maps are homeomorphisms

why we need quotient? Would not the theorem I stated above make the solution simpler? Or you want the quotient to implement the first comment below the other answer below?

Also, in your map, why you took any two points in the first $D^n$ while specifically the points 0 and 1 in the second $D^n$?

@Intuition We need a quotient map because pushouts are constructed as quotients! To say $S^n$ is (homeomorphic to) the pushout of $D^n\leftarrow S^{n-1}\rightarrow D^n$ is precisely to say there is a quotient map $q:D^n\sqcup D^n\twoheadrightarrow S^n$ with certain properties. In this case, those properties boil down: $q(x)=q(y)$ iff. $x,y$ represent the same point on the boundary spheres $S^{n-1}\subset D^n$. I use $(x,0)$ and $(y,1)$ because that's a standard way to distinguish between the copies of $D^n$ - $D^n\sqcup D^n=D^n\times\{0\}\cup D^n\times\{1\}$ by some definitions.

so you are saying that my theorem can not be used here, did I understand you correctly ?

do you know where exactly is this quotient theorem? Either in Munkres or Allen Hatcher, point set topology?

@Intuition If you really want: I gave a continuous map $D^n\sqcup D^n\to S^n$ that agreed on the common subsphere $S^{n-1}$. That induces a continuous map $D^n\cup_{S^{n-1}}D^n\to S^n$. My map was surjective so this map is also surjective. If you know how to construct pushouts you can also check this map is injective. So it's a continuous bijection, and because of how pushouts are constructed you realise pushouts of compact spaces are compact so it follows this map is a homeomorphism. But that's exactly the same as what I did, except what I did is shorter and more conceptual

the explanation I just gave where we literally use the "continuous bijection = homeomorphism" theorem implicitly uses properties of quotient spaces. You'd be better off applying those properties of quotient spaces directly to see why my map makes $S^n\cong D^n\cup_{S^{n-1}}D^n$ as explained in the post. The quotient theorems should be in Munkres; it's almost certainly not in Hatcher. It's definitely in Brown's 'topology and groupoids'

Do you have a nicer explanation to this solution math.stackexchange.com/questions/4205392/…

@Intuition Yes, but also no. Henno's answer is already great. The right thing to do is ask a new question inquiring about motivation and further details. There is nothing wrong with Henno's answer

I am not saying it is wrong at all, in my clarification for the bounty I said exactly the reasons for the bounty ... You can take a look at it and let me know your thoughts again please.

@Intuition I know the reasons for the bounty. It just feels disrespectful to write a new answer when the original answer is already complete and I don't exactly have much to add. I did it in this case because the original answer was lacking rigorous detail; for this intuition question, since it is essentially a different question ! I encourage you to ask it in a new post.

ok I see your point and I respect it.

So, basically here in your proof , to prove the required you are using the idea that injective quotient maps are homeomorphisms .... Am I correct?

@Intuition that’s one way to think about it but in my opinion it’s not the best way. Really, the point is that quotient spaces have a universal property, and one consequence of this is the following theorem: if $q:X\to Y$ and $q’:X\to Y’$ are quotient maps inducing the same relation on $X$, meaning $q(x_1)=q(x_2)\iff q’(x_1)=q’(x_2)$, then there is a unique homeomorphism $\phi:Y\cong Y’$ such that $\phi q=q’$. In particular $Y$ is homeomorphic to $Y’$. Since there is a quotient map $D^n\sqcup D^n\to S^n$ inducing the correct relation, then there is a homeomorphism with the pushout for free.

It is more intuitive for me that the point $(x,0)$ goes to the negative hemisphere and $(y,1)$ goes to the positive hemisphere, what do you think?

@Intuition That's also fine. It doesn't really matter. In fact the choice of hemispheres is arbitrary, I could rotate. It just keeps the explicit formulas a lot nicer this way. But as you say, $(x,0)\mapsto(x_1,\cdots,x_n,-\sqrt{1-\|x\|^2})$ and $(y,1)\mapsto(y_1,\cdots,y_n,\sqrt{1-\|y\|^2})$ works too

where does the points $(z,0)$ and $(z,1)$ lies? Do they lie on the equator? What does it mean for z to has norm 1?it means we are on the boundary of the disk, right?

@Intuition Yes, norm one means it is on the boundary of the disk, the subsphere $S^{n-1}$. They do indeed lie on the equator, as the final component $\sqrt{1-\|z\|^2}=\sqrt{1-1^2}=\sqrt{0}=0$

I believe the equivalence relation to get the pushout is just $i(a) = i(a)$ for every a in $S^{n-1}$ but this is trivially correct, am I right? So, what is the equivalence relation of our quotient here? ..... I believe our given two disks are separated by one so how this is reflected in the equivalence relation?

did you mean in the paragraph before last to say embedding $S^{n-1}$ into $D^n$? Not $S^n$?

@Intuition $S^n$ does not embed into $D^n$. This is not a trivial fact but hey. I definitely meant $S^{n-1}$! Remember $S^{n-1}$ is defined as a subset of $\Bbb R^n$, as is $D^n$... the indexing is easy to mess up but $S^{n-1}\subset D^n$ is right

what about my comment before this one?

@Intuition Ah, so remember we are dealing with two different copies of $D^n$. Really you should ask for the equivalence relation generated by $i_0(a)=i_1(a)$, where $i_e:S^{n-1}\hookrightarrow D^n\times\{e\}$, $e=0,1$. Picture two different 2-discs and laying them on top of one another, gluing at the circumference. That’s what we want to do; the disks are different ! except at the boundary where we force them to be the same

why the given map after quotienting by this equivalence relation is still well defined?

@Intuition Well, I discuss this in my answer. My map when composed with $i_1$ and $i_0$ gives the same composite function, which is all we need. The final coordinate becomes zero and $0=-0$, and the maps are the same on all the other coordinates.