Does a homeomorphism of open cones restrict to a quotient map of the bases?

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.

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