Theorem: Let $M$ be an $n-$manifold with orientation $\mu$. Then for each compact $K \subset M$ there exists a unique class $\mu_K \in H_n(M,M-K)$ such that $$\rho_x(\mu_K) = \mu_x$$ for each $x \in K$.
The uniqueness is not a problem. I don't understand the existence, in particular the "uniformity" over $K$.
My question regards the case of $K \subset U$ where $U$ is open homeomorphic to $\mathbb{R}^n$, which should the equivalent of the non-treated case of Massey, citing: "Obviously if the compact is contained in a sufficiently small neighborhood of some point, the continuity condition of $\mu$ assures us the existence of $\mu_K$".
I tried the following: take a disk $D_R^n$ big enough such that $K \subset D_R^n$ and consider the commutative diagram
$$\begin{array}{cc} \,\,\,\,\,\,\,\mathbb{Z} = H_n(\mathbb{R}^n,\mathbb{R}^n \setminus D_R^n) \\ \\ \varphi \downarrow & \,\,\,\,\,\,\,\searrow \,{\simeq} \\ \\ \,\,\,\,\,\,H_n(\mathbb{R}^n,\mathbb{R}^n \setminus K) & \xrightarrow{\rho_x} & \mathbb{Z} = H_n(\mathbb{R}^n,\mathbb{R}^n \setminus \left\lbrace x\right\rbrace) \end{array}$$
Then we can take $\mu$ generator of $H_n(\mathbb{R}^n,\mathbb{R}^n \setminus D_R^n)$ that under the isomorphism on the right(given by excision) goes to $\mu_x$. This allow to define $\mu_K := \varphi(\mu)$, in this way clearly $\rho_x(\mu_K) = \mu_x$.
This proves the assertion only for the particular $x$ chosen, and not for all $x \in K$, so my proof is stuck here. I don't know that if in order to prove the thesis for all $x \in K$ I should use the following observation :
If we take $x,y \in M$ that lay in the same ball $D^n \subset M$ then we have
$$ \mathbb{Z} = H_n(M,M- \left\lbrace x \right\rbrace) \leftarrow \mathbb{Z} = H_n(M,M- D) \rightarrow H_n(M,M- \left\lbrace y \right\rbrace)$$
And the map are isomorphism, so we have a canonical isomorphism between $H_n(M,M- \left\lbrace x \right\rbrace)$ and $H_n(M,M- \left\lbrace y \right\rbrace)$.
Any help to finish my proof or any other proof would be appreciated, thanks in advance.
Say $M=\mathbb R^n\newcommand\R{\mathbb R}\newcommand\sm\setminus$, and fix a compact subset $K\subset\mathbb R^n$. For each $x\in K$, there exists a small open $U_x$ (edit: say, $U_x$ is literally an open ball) around $x$ and a generator $\mu_{U_x}\in H_n(\R^n,\R^n\setminus U_x)$ so that $\mu_{U_x}$ maps to the chosen generators $\mu_y\in H_n(\R^n,\R^n\sm\{y\})$ for all $y\in U_x$.
Since $K$ is compact, finitely many such $U_x$'s will do, so let $U_1,\dots,U_N$ be a finite open cover of $K$ with generators $\mu_i\in H_n(\R^n,\R^n\sm U_i)$ mapping to the preferred generators in $H_n(\R^n,\R^n\sm\{x\})$ for all $x\in K$. We want to do some induction shortly, but let's first set up a bit of notation. For any $k\in\{2,3,\dots,N\}$, let $$V_k:=\bigcup_{i=1}^kU_i\text{ and }W_k:=U_k\cap V_{k-1}=\bigcup_{i=1}^{k-1}(U_k\cap U_i).$$ Then, relative Mayer-Vietoris gives us an exact sequence $$H_n(\R^n,\R^n\sm V_k)\to H_n(\R^n,\R^n\sm V_{k-1})\oplus H_n(\R^n,\R^n\sm U_k)\to H_n(\R^n,W_k)$$ where the first map is what you expect and the second map is $(a,b)\mapsto a-b$. Inductively assume that we have some generator $\nu_{k-1}\in H_n(\R^n,\R^n\sm V_{k-1})$ so that $\nu_{k-1}\mapsto\mu_x\in H_n(\R^n,\R^n\sm\{x\})$ for all $x\in V_{k-1}$ (note $\nu_1=\mu_1$). Then, $(\nu_{k-1},\mu_k)\mapsto0$ under the second map in our exact sequence, so is the image of a unique $\nu_k\in H_n(\R^n,\R^n\sm V_k)$.
The upshot is that Mayer-Vietoris allows us to glue our original $\mu_i$'s into one $\nu:=\nu_N\in H_n(\R^n,\R^n\sm V_N)$ Since $K\subset V_N$ by construction, we can let $\mu_K$ be the image of $\nu$ under $H_n(\R^n,\R^n\sm V_N)\to H_n(\R^n,\R^n\sm K)$.