I am thinking about the equivalence relation of the pushout here $D^n \longleftarrow S^{n-1} \longrightarrow D^n,$ is it just $i(a) = i(a)$ for all $a \in S^{n-1}$ where $i$ is the inclusion map $S^{n-1} \longrightarrow D^n,$ but this is trivially true or I am missing something? I am trying to understand exactly what quotient do we have.
Any explanation will be greatly appreciated!
Abstractly, the pushout is a space $P$ together with $f_1,f_2\colon D^n\to P$ (one for each copy of $D^n$), such that the corresponding diagram commutes, i.e., the two maps $f_k\circ i_k\colon S^{n-1}\to P$ are the same. Moreover, for any other space $X$ together with $g_1,g_2\colon D^n\to X$ and such that $g_1\circ i_1=g_2\circ i_2$, there exists a unique $h\colon P\to X$ such that ... yada yada.
In other words, $f_1$ and $f_2$ can be expected to agree only on the boundary, whereas they must be able to accommodate arbitrarily differing $g_1,g_2$ at all other points of the $D^n$'s. But this just boils down taking two copies of $D^n$ and glueing them together by identifying their boundaries $S^{n-1}$. The result can be viewed as $S^{n}$, with the given $S^{n-1}$ being the "equator" and the two copies of $D^n$ being the two closed "hemispheres".