Write $CX$ for the (pointed, or reduced) cone on $X$, and $C^\circ X$ for the open cone inside of it.

Let's say a cone map is a map $g:CX\to CY$ such that $g(C^\circ X) \subseteq C^\circ Y$ and $g(X) \subseteq g(Y)$.

Let $A$ and $B$ be two cell complexes -- that is, spaces built from $*$ by iteratively attaching wedges of disks by various attaching maps, but not necessarily in dimensional order. Suppose $f:CA\to CB$ is a cone map; then $f$ restricts to two maps $$ f_{\mathrm{base}}: A \to B \qquad \mbox{and} \qquad f^\circ : C^\circ A \to C^\circ B. $$

QUESTION: If $f^\circ$ is a homeomorphism, then must $f_{\mathrm{base}}$ be a quotient map?

NOTE 1: The map $f_{\mathrm{base}}$ need not be a homeomorphism. For example, let $f:D^2\to D^2$ be a map that collapses a small segment of the boundary to a point.

NOTE 2: I would be interested to see answers for unpointed cones, even though my primary reason for asking this question has to do with the pointed case.

NOTE 3: $q:X\to Y$ is a quotient map if $t:Y\to T$ is continuous if and only if $t\circ q$ is continuous; I don't insist that quotient maps be surjective, but I believe the maps in question here are surjective anyway.


I'll construct a counterexample with reduced cones, which restricts to the map $f$ from https://math.stackexchange.com/a/415666/727733 on the base.

Let $X=C[0,2\pi)=[0,2\pi)\wedge [0,1]$ (with special points 0). Without $[0,2\pi)\times \{1\}$ this is homeomorphic to the open disk. Choose an appropriate homeomorphism and extend it to $[0,2\pi)\times \{1\}$, sending it to the boundary of the disk by $f$. As the closed disk can be thought of as the cone above a circle, this is indeed a counterexample.

  • $\begingroup$ Very nice! Using the reduced cones allows you to eliminate the difference in topology between the half-open interval and the circle. $\endgroup$ – Jeff Strom Nov 25 at 15:40

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