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?

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    $\begingroup$ 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. $\endgroup$ – William Dec 11 '19 at 4:36
  • $\begingroup$ @William You should give an official answer. $\endgroup$ – Paul Frost Dec 11 '19 at 16:10

As it is written your statement is not correct, here is a counter-example:

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.

  • $\begingroup$ I got it! Thanks a lot! $\endgroup$ – HooMun Dec 12 '19 at 1:31

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