Let $$F: (X,A,*)\times [0,1]\to (X,A,*)$$ be a homotopy such that $F_0$ is the identity and $F_1$ maps $A$ to the point $*$. Does it imply that $X$ is homotopy equivalent to $X/A$? If it helps, we may assume that all spaces are $CW$ complexes or simplicial sets.
I tried to analyze the maps $\tilde{F_1}: X/A\to X$ induced by $F_1: (X,A)\to (X,*)$ and possible homotopy inverse $q: X\to X/A$ but am not able to prove that they induce homotopy equivalence.
No, it's not always true that $X/A \simeq X$ when $A$ is contractible. It is true if the pair $(X,A)$ has the homotopy extension property, as proven in Hatcher's book as Proposition 0.17.
Here is a counterexample. Let $X = S^1$ be the circle, let $x \in X$ be an arbitrary point, and let $A = S^1 \setminus \{x\}$. Then $X/A$ has exactly two elements, say $x$ and $y = [A]$, and its open sets are $\varnothing$, $\{y\}$ and $X/A = \{x,y\}$. This is a contractible space and hence not homotopy equivalent to $X = S^1$.
