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.



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


