What will be the pushout for the following :
where $i:S^{n-1} \rightarrow D^n$ is the inclusion of the boundary $S^{n-1}$ to the n-disk $D^n$.
According to Pg 40 in Julia E. Bergner's The Homotopy Theory of (∞,1)-Categories,
pushout space should be $S^n$ and I am guessing quotient maps $\pi:D^n \rightarrow D^n/S^{n-1}=S^n$ will be its projection maps. I was trying to use the Universal property of the quotient map to prove the Universal property of Pushout.
My attempt:
Let $\phi,\psi:D^n \rightarrow X$ be two continuous maps to some topological space $X$ such that $\psi \circ i= \phi \circ i$
But to use the Universal property of quotient map or in other words to produce a unique $\theta: S^n \rightarrow X$ we need $\phi$ and $\psi$ both have to be constant and equal on $S^{n-1}$. But I am not seeing any reason why any arbitrary map $D^n \rightarrow X$ will be constant on the boundary.
So are my choices of projection maps $(\pi, \pi)$ are wrong ones? Or am I misunderstanding something?
I am confused.
Thanks in advance.
You must not take the quotient map $\pi : D^n \to D^n/S^{n-1}$.
Instead let $j_+ : D^n \to S^n, j_+(x) = (x,\sqrt{1 - \lVert x \rVert^2})$, and $j_- : D^n \to S^n, j_+(x) = (x,-\sqrt{1 - \lVert x \rVert^2})$. These maps embed $D^n$ as the upper and lower closed hemisphere. Then $\require{AMScd}$ \begin{CD} S^{n-1} @>{i}>> D^n \\ @V{i}VV @V{j_+}VV \\ D^n @>{j_-}>> S^n\end{CD} is a pushout diagram. Im am sure you can fill the necessary details.
By the way, the canonical construction of the pushout is to take the disjoint union $D^n + D^n$ of two copies of $D^n$ and to identify the copies of $S^{n-1} \subset D^n$ via the identity map. This clearly gives a space homeomorphic to $S^n$.