Quotient map preserves deformation retraction

For the statement:

Suppose the subspace $$A$$ is a deformation retract of the space $$X$$, and the quotient map $$q: X \rightarrow Y$$ is given. Then, $$q(A)$$ is deformation retract of the space $$Y$$.

My attempt starts with defining a continuous function $$H$$ on $$X \times I$$ to $$X$$ satisfying $$H(x, 0)=x,\ H(x, 1) \in A,\ H(a, t)=a$$ for any $$x \in X, a \in A$$, and $$t\in I$$.

Since the composite $$q\circ H$$ is continuous, and $$q: X \rightarrow Y$$ is a quotient map, using the universal property, I want to define a continuous map $$G: Y \times I \rightarrow Y$$ that gives a deformation retraction between $$Y$$ and $$q(A)$$. However, $$q \circ H$$ is not constant on $$q^{-1} (z)$$ for each $$z \in Y$$, because the pre-image of $$z$$ could lie in $$X-A$$. Is this approach right to solve the problem?

• I don't think this is true as stated. Take $X = [0,1]$, $A = \{0\}$, $Y = S^1$, and $q(t) = e^{t * 2\pi i }$, i.e. $q$ is the quotient map that identifies $0$ and $1$. Then although $\{0\}$ is a deformation retract of $[0,1]$, $q(\{0\})$ is not a deformation retract of $S^1$ because $S^1$ is not contractible. – William Dec 11 '19 at 4:36
• @William You should give an official answer. – Paul Frost Dec 11 '19 at 16:10

Take $$X = [0,1]$$, which deformation retracts onto $$A = \{0\}$$. Now consider the quotient $$Y = X/(0 \sim 1)$$, which is homeomorphic to $$S^1$$. The quotient map $$q \colon X \to Y$$ takes the singleton $$\{0\}$$ to the single equivalence class containing $$0$$ and $$1$$, but $$Y$$ cannot deformation retract onto this point because $$Y\cong S^1$$ is not contractible.