Two disks glued by any boundary homeomorphism form a topological sphere

Let $m\ge0$, let $D^{m+1}=\{x\in\mathbb R^{m+1}:\|x\|\le1\}$, and let $S^m=\partial D^{m+1}$ carry its usual subspace topology. For any homeomorphism $\varphi:S^m\to S^m$, form the quotient of two labelled copies of the disk by the equivalence relation generated by identifying the left boundary point $u$ with the right boundary point $\varphi(u)$. With the quotient topology,

Here $S^{m+1}$ is the standard unit sphere in $\mathbb R^{m+2}$, and $\cong$ denotes existence of a homeomorphism. There is no orientation or differentiability assumption on $\varphi$; the assertion includes $m=0$.

This is a uniform topological gluing result for an explicitly specified pair of standard disks. At $m=3$ it identifies that quotient with the topological four-sphere. It does not assert that an arbitrary homotopy four-sphere admits such a disk decomposition, and it neither proves Freedman's general recognition theorem nor establishes smooth standardness or the smooth Poincaré conjecture.