Cofiber of a based cofibration is equivalent to the quotient

Let r retract the reduced cylinder on X to $M_i=X{\cup }_i(A\times I)$ with basepoint track collapsed. Collapse $A\times \{1\}$ in the target to obtain $R:X\times I\to C_i$. It satisfies $R(x,0)=x$ and $R(a,t)=[a,t]$. In particular $R(a,1)=\ast$, so $R(-,1)$ descends to a based map $\varphi :X\mathrm{/}A\to C_i$. All maps are continuous through their quotient topologies by F2 and F3.

The homotopy $\psi R(x,t)$ is constant on A for every t, since R(a,t) lies in CA. It is constant on the fibres of $(X\to X\mathrm{/}A)\times {\mathrm{id}}_I$, so F3 and F4 descend it to a homotopy on X/A. At t=0 it is the identity and at t=1 it is $\psi \varphi$.

On X use R(x,t), and on the attached cone use $[a,s]\mapsto [a,\max (s,t)]$. At s=0 these agree by step 1.1; at s=1 the image is always the tip, and on the basepoint track it is always *. These compatible formulas descend through the cofiber quotient and its product with I by F3 and F4. At t=0 the resulting homotopy is the identity of $C_i$; at t=1 it agrees with $\varphi \psi$ on X and sends CA to *. Along with step 2.1 this proves the based homotopy equivalence.