If $A$ is a path connected subspace, is $q_*: \pi_1(X) \to \pi_1(X/A)$ a surjection?
It's not surjective in general. Consider the topologist sine curve $S$
$$S = \left\{\left(x, \sin \bigg(\frac{1}{x}\bigg)\right),\ 1/\pi\geq x>0\right\}$$
together with the interval $A = \{(0, t): |t|\leq 1\}$ and a curve $C$ joining this interval to the point $(1/\pi, 0)$. Call $X = A \cup S\cup C$. Then this is simply connected and $X/A$ is homeomorphic to $\mathbb S^1$, thus the induced map is zero.
Remark: To see that $X/A$ is homeomorphic to $\mathbb S^1$ Let $L = \{ (t,0) : t\in [0,1]\}$. Then $\mathbb S^1$ is homeomorphic to $B:=L\cup C$. Define $ F : X\to B$ by
$$F(x_1, x_2) = \begin{cases} (x_1,0) & \text{if } (x_1, x_2) \in A\cup S, \\ (x_1, x_2) & \text{if } (x_1, x_2) \in C. \end{cases}$$
One can check that $F$ is continuous and $f(A) = \{ (0,0)\}$. Thus $F$ descends to a continuous mapping $f: X/A \to \mathbb S^1$ which is bijective. Since $X/A$ is compact (Since $X$ is) and $\mathbb S^1$ is Hausdorff, $f$ is a homeomorphism.
For CW complexes, you can do the following:
- Use the long exact sequence of homotopy groups to prove that $\pi_1(X)\rightarrow \pi_1(X,A)$ is surjective.
- Use excision for homotopy groups (Hatcher, Prop. 4.28) to prove that $\pi_1(X,A)\rightarrow \pi_1(X/A)$ is surjective.
EDIT: There is a shorter argument using Van Kampen's theorem: $X/A$ is homotopy equivalent to the union of $X$ and $CA$ (the cone of $A$) along $A$. As all three spaces are path connected, we have that the homomorphism $\pi_1(X)*\pi_1(CA)\to \pi_1(X/A)$ is surjective, and since $CA$ is contractible, we have that $\pi_1(X)\to \pi_1(X/A)$ is surjective as well. Notice that this works in more generality than CW-complexes.