Proof that a certain section of the orientation cover is continuous I am wondering whether I on the right track to show continuity of a certain function. Here is some background:
Let $M$ be an $n$-manifold. Given a subset $A \subset M$ and $\alpha \in H_n(M,M\backslash A)$, define $\alpha_x$ to be the image of $\alpha$ under the map (induced by inclusion of pairs) $H_n(M,M\backslash A) \to H_n(M,M\backslash x)$. Hatcher defines a covering space $M_\mathbb{Z}$ of $M$, where $M_\mathbb{Z} = \coprod_{x\in X} H_n(M,M\backslash x)$, and the covering map sends all of $H_n(M,M\backslash x)$ to $x$. A basis for the topology on $M_\mathbb{Z}$ is given as follows: for a ball $B \subset M$ and $\beta \in H_n(M,M\backslash B)$, the subset $U(\beta) = \bigcup_{x\in B} \beta_x$ is declared to be open. 
Hatcher mentions in the discussion after Lemma 3.27 that given a compact subset $A\subset M$ and an element $\alpha \in H_n(M,M\backslash A)$, the section (as sets) $\widehat{\alpha} : x\mapsto \alpha_x \in H_n(M,M\backslash x)$ is continuous. 
Here is my proof: 

Since the $U(\beta)$ are a basis for the topology on $M_\mathbb{Z}$, we just need to show that $\widehat{\alpha}^{-1}(U(\beta))$ is open, where $\beta \in H_n(M,M\backslash B)$ and $B$ is a ball in $M$ (which we may as well assume intersects $A$, otherwise the preimage is empty). Since $\widehat{\alpha}$ and $\widehat{\beta}$ are sections as set functions at least, we have $\widehat{\alpha}^{-1}(U(\beta)) \subset A\cap B$.  Since $\widehat{\alpha}$ and $\widehat{\beta}$ are both lifts of the inclusion map $A\cap B \hookrightarrow M$, $\dots$

If $A\cap B$ were connected for some reason, I would like to argue that $\widehat{\alpha} $ and $\widehat{\beta}$ must either agree on all of $A\cap B$, or on none of it. This seems unlikely for general subsets. Perhaps this is where we use compactness of $A$?
Am I on the right track here? It seems like I'm making things too difficult. 
 A: I will use the notation $f$ instead of $\hat {\alpha}$ just for convenience. We have  to show that the section $f:x \mapsto \alpha_x$ is continuous. Fix a point $x \in M$. It suffices to show that $f$ is continuous at $x$. Let $U:=U(\beta_B)$ be any basis open set of $M_\Bbb Z$ containing the point $f(x)=\alpha_x$. We have to show that $f^{-1}(U)$ contains a neighborhood of $x$ in $M$. (Here, the notation $U(\beta_B)$ is as in Hatcher's. Thus, $B$ is an open ball such that $x \in B \subset  \Bbb R^n \subset M$ and $\beta_B \in H_n (M,M-B;R)$. Also, we are identifying a neighborhood of $x$ in $M$ that is homeomorphic to $\Bbb R^n$ with $\Bbb R^n$, again as in Hatcher's. I think you might be familiar with these because you were reading Hatcher's.). 
We will in fact show that $f^{-1}(U)=B$. We obviously have the inclusion "$\subset$". Note that, since $\alpha_x \in U$, we have $\beta_B \mapsto \alpha_x$ under the map $H_n(M,M-B;R) \to H_n(M,M-x;R)$, by the definition of $U$. On the other hand, under the map $H_n(M;R) \to H_n(M,M-x;R)$ we have $\alpha \mapsto \alpha_x$. Since this map factors through $H_n(M,M-B;R)$, we must have $\alpha \mapsto \beta_B$ under the map $H_n(M;R) \to  H_n(M,M-B;R)$ (because $H_n(M,M-B;R) \to H_n(M,M-x;R)$ is an isomorphism, and in particular injective).
Finally, to show that $B \subset f^{-1}(U)$, let $y \in B$. By hypothesis we have $\alpha \mapsto \alpha_y$ under the map $H_n(M;R) \to H_n(M,M-y;R)$, which factors through $H_n(M,M-B;R)$. Hence, by commutativity, we must have $\beta_B \mapsto \alpha_y$ under the map $H_n(M,M-B;R) \to H_n(M,M-y;R)$, which implies $f(y)=\alpha_y \in U$, or equivalently $y \in f^{-1}(U)$, as desired, and the proof is complete.
The proof seems long because I wrote all the details, but in fact there are no difficult ideas, i.e., this is a quite straightforward argument.
