Assume $(X,A)$ has the homotopy extension property (or even a CW pair). Suppose we are given a map $f : X \to Y$, such that the restriction $f|_A : A \to Y$ is nullhomotopic for some subspace $A \subset X$. Then, given a nullhomotopy $h : A \times I \to Y$, we can solve the homotopy extension problem and get a map $F : X \times I \to Y$ such that $$F|_{X\times 0} = f, \; F|_{A\times I} = h$$ In particular, $F|_{A\times 1} = h|_{A\times 1} = *$, and we have an induced map $g : X/A \to Y$.
Question: Does the map $g$ depends (up to homotopy) on the choice of the nullhomotopy?
Here is the example that I had in mind. Consider, $(X, A) = (D^2, S^1)$. Consider the map $f : D^2 \to S^2$ which maps identically onto the upper hemisphere. Now, $f|_{S^1}$ is clearly nullhomtopic, since $\pi_1(S^2)=0$. Denote by $h_u : S^1 \times I \to S^2$ the homotopy which slides the meridian circle over the upper hemisphere and to a point on the meridian (while keeping this point fixed). Denote by $h_l : S^1 \times I \to S^2$ the homtopy which slides along the lower hemisphere.
Then, it seems to me, that $h_u$ induces the constant map $D^2/S^1 \to S^2$, whereas $h_l$ induces the quotient map $D^2/S^1 \to S^2$. Clearly, these are not homotopic, since the quotient map is really the identity $S^2 \to S^2$.
Is my (non)example correct, or am I making some silly continuity mistake?! Any help is appreciated. Cheers!