I was working on the following exercise (1.9.28 from Shastri's Basic Algebraic Topology), where $SX$ is the unreduced suspension of a topological space $X$:
I'm having problems figuring out a way to prove point $(g)$: I thought about the fact that given a cofibration $A\hookrightarrow Y$ with $A$ contractible then the map $Y\rightarrow Y/A$ is a homotopy equivalence. The cone $CX$ is contractible, but does it follow from $\{x_0\}\hookrightarrow X$ being a cofibration that $CX\hookrightarrow SX$ is such?
Another possibility was that the cofibration $\{x_0\}\rightarrow X$ induces a retraction $X\times I\rightarrow \{x_0\}\times I \cup X\times\{0\}$ which can be used to build a retraction $CX\rightarrow \{x_0\}\times I$, but is this useful somehow? ($I=[0,1]$)
Finally, in point $(f)$ it is asked to build a homeomorphism $(SX,C_-X)\rightarrow(C_+X,X)$: can this really be done or is it just a continuous map of topological pairs? The demand that $C_-X$ be sent inside $X$ seems way strong to me
Edit: as user87690 pointed out in the comments, being a relative homeomorphism for a topological pair map $(X,A)\rightarrow(Y,B)$ implies that it is a continuous map of topological pairs mapping homeomorphically $X- A$ onto $Y-B$: hence the map fixing the upper cone and projecting the lower cone onto the "horizon" is such, aswering to point $(f)$