A compact convex set $A$ is contained in $\mathbb{R}^n$ and $R$ is a ring with an identity. I need to show that for any section $$x \rightarrow a_x \in H_n(\mathbb{R}^n|x ; R)$$ there is a unique $a_A \in H_n(\mathbb{R}^n|A;R)$ such that for each $x \in A$, $$a_A \rightarrow a_x$$ in the carnonical map $$H_n(\mathbb{R}^n|A;R) \rightarrow H_n(\mathbb{R}^n|x;R)$$.
This is from the proof of Lemma 3.27 in Hatcher's book, from which it states that this is obvious since $H_n(\mathbb{R}^n|A;R) \rightarrow H_n(\mathbb{R}^n|x;R)$ is an isomorphism. It's easy to see that the map is an isomorphism but I find it difficult to point out the existence of consistent $a_A$.
Every section determine an orientation. The space $M^*=\{ a_x|x \in \mathbb{R}^n \& a_x \in H_n(\mathbb{R}^n|x ; R) \}$ consists of $M_r = \{a_x|a_x = \pm r \}$ for each $r \in R$ and $M_r = M_{-r}$ and with the compact of path it's easy to show that for $r_1 \neq \pm r_2$, $M_{r_1}$ and $M_{r_2}$ is path disconnected. So for an section $$s: x \rightarrow a_x$$ we have $a_x = \pm r$ for some $r$. If $r = -r$ this is obviously an orientation or otherwise we can derive an orientation from this section that is $$x \rightarrow \begin{cases} \mu_x & \text{if } s(x) = r,\\ -\mu_x & \text{if } s(x) = -r. \end{cases}$$ where $\mu_x$ is the generator of $H_n(\mathbb{R}^n | x ; \mathbb{Z})$ when regarding $H_n( \mathbb{R}^n | x ; R)$ as tensor product $H_n(\mathbb{R}^n | x ; \mathbb{Z}) \otimes R$. This can be solved with basic topological method.
So every section is a homeomorphism from $\mathbb{R}^n$ to a component of $M_r$. And there is an open ball $B$ contains $A$. $H_n(\mathbb{R}^n|B) \rightarrow H_n(\mathbb{R}^n |x_0)$ is an isomorphism where $x_0 \in A$ so there is $a_B \rightarrow a_{x_0}$. And $U_B = \{ a_x | x \in A \& a_B \rightarrow a_x \}$ is which commonly defined a topology i.e. connected open set intersecting the image of the section and hence contained in the image of the section. $H_n(\mathbb{R}^n|B) \rightarrow H_n(\mathbb{R}^n |x_0)$ factors through $H_n(\mathbb{R}^n | A)$ and $H_n(\mathbb{R}^n|B) \approx H_n(\mathbb{R}^n |x_0) \approx H_n(\mathbb{R}^n | A)$ so we have $a_B \rightarrow a_A \rightarrow x$ for all $x \in A$.
Though from preceding I solved it much naturally, it seems from the tone of the book that there is a more direct way to point it out, which is from the fact that $H_n(\mathbb{R}^n |A) \rightarrow H_n(\mathbb{R}^n | x)$ is an isomorphism.