What is the pushout of $D^n \longleftarrow S^{n-1} \longrightarrow D^n$?

Question

I am trying to prove that the pushout of $D^n \longleftarrow S^{n-1} \longrightarrow D^n$ is homeomorphic to $S^n$. How should I progress?

Answer 1

It is just the sphere $S^n$ obtained by gluing two $n$ discs by their boundaries.

Answer 2

In response to the bounty request.

We seek a suitable quotient map $D^n\sqcup D^n\to S^n$. Map: $$\begin{align}(x,0)&\mapsto(x_1,x_2,\cdots,x_n,\sqrt{1-\|x\|^2})\quad\text{ on the first copy of $D^n$}\\(y,1)&\mapsto(y_1,y_2,\cdots,y_n,-\sqrt{1-\|y\|^2})\quad\text{ on the second copy of $D^n$}\end{align}$$This is obviously well defined and continuous; it is a map of compact Hausdorff spaces, so if it is surjective then it is a quotient map. And indeed it’s easy to see it is surjective, so we realise $S^n$ is some quotient of $D^n\sqcup D^n$.

Now we just need to check it’s the correct quotient. What are the fibres? Evidently this map injects for points with norm strictly less than one (no $(x,0)$ can be identified with any $(y,1)$ in these cases as the signs of their last coordinates are positive, negative) and clearly it identifies $(z,0)\sim(z,1)$ exactly when $z$ has norm one; by looking at the first $n$ coordinates you see no other identifications could be made. You can convince yourself it really does enforce the quotient relation associated to $S^n\hookrightarrow D^n$. Done!

This map is extremely geometric too. It is the northern/southern hemisphere decomposition; projection onto the first $n$ coordinates corresponds to pushing the sphere down to its equator; all we are doing is taking a disk and lifting its points, adjusting only their heights, suitably so it rests on the sphere’s surface.