I have the following qual problem:
Let $M$ be a connected closed surface, not necessarily orientable, with an embedded closed disk $D$. Let $Q$ be the quotient space of $M$ by $\overline{M\setminus D}$. After showing $Q\cong S^2$ show that the quotient map $q: M \rightarrow S^2$ is not nullhomotopic.
If $M$ is orientable, I compute the degree of $q$ to be non-zero and deduce that the map is not nullhomotopic. However, if $M$ is non-orientable, then I cannot think of what to do. Since both the quotient map and a constant map descend to the quotient space $Q$, I would like to modify a nullhomotopy to descend to $Q$ and then arrive at a contradiction, but do not see how to pull that off. I have also thought about using, somehow, the oriented double cover of $M$.
Consider the composite map $$(D.\partial D) \rightarrow (M,M-D) \rightarrow (S^2, x_0)$$ It suffices to show that $(D.\partial D) \rightarrow (S^2, x_0)$ is not nullhomotopic. For example it induces an isomorphism in homology $H_2(D.\partial D) \rightarrow H_2(S^2, x_0)$.