Is local homology group on a manifold a sheaf? Let $X$ be a manifold of dimension $n$, and define $\mathcal{F}(U) = H_n(X,X-U)$. Then clearly $\mathcal{F}$ is a presheaf. I am thinking whether $\mathcal{F}$ is a sheaf. 
According to Lemma 3.27 in Hatcher's algebraic topology, $\mathcal{F}$ satisfies identity and gluability axiom on compact sets $A$. So I am guessing $\mathcal{F}$ is not a sheaf in general.
Is this true? What conditions do $X$ need to satisfy such that $\mathcal{F}$ is a separated presheaf, and a sheaf?
 A: $\mathcal{F_k}$ defined by $\mathcal{F_k}(U) = H_k(X, X-U;R)$ for $R$ a ring is not a sheaf. Consider for example $X = \mathbb R^n$, $R = \mathbb Z$, and $k = n$: if $B_r(0)$ denotes the open ball of radius $r$ around zero, then $\mathcal{F}(B_r(0)) \cong \mathbb{Z}$ for all $r > 0$, but compatible classes in $\mathbb{F}(B_r(0))$ cannot be glued, as $\mathcal{F}(\mathbb R^n) = H_n(\mathbb R^n; R) = 0$. In fact, this example is general, as every $n$-manifold is covered by neighborhoods homeomorphic to $\mathbb R^n$.
But, (when using singular homology,) $\mathcal{F}_k$ is a separated presheaf for all $k$, which can be used to show that $\mathcal{F}_k = \underline{0}$ for $k > n$ and that $\mathcal{F}_n$ satisfies finite gluing.
Separation: suppose that $U = \cup_\alpha U_\alpha$ and $[z] \in \mathcal{F_k}(U)$ satisfies that $[z]|_{U_\alpha} = 0$ for all $\alpha$, where $z$ is a representing chain. Then for all $\alpha$, $z = \partial y_\alpha$ for some $y_\alpha \in C_{k+1}(X, X-U_\alpha; R),$ which implies that the union of the images of the nonzero simplices in $z$ is contained in $X - U_\alpha$ for all $\alpha$. Hence, the image is contained in $\cap_\alpha (X - U_\alpha) = X - U$, so any such $y_\alpha$ defines a chain in $C_{k+1}(X, X-U;R)$. Hence, $[z] = 0$.
$\mathcal F_k = \underline{0}$ for $k > n$: since $X$ is an $n$-manifold, $X$ has a basis of balls $B_\alpha$ contained in open sets $U_\alpha \cong \mathbb R^n$. By excision, $\mathcal F_k(U_\alpha) \cong H_k(U_\alpha, U_\alpha - B_\alpha) = 0$.
Since the presheaf is separated and is zero on a basis of open sets, the presheaf is identically zero.
$\mathcal F_n$ satisfies finite gluing: Say that $U_1$ and $U_2$ are open sets. Then the Mayer-Vietoris sequence for homology gives us the exact sequence
$$ 0 \to H_n(X, X - (U_1 \cup U_2);R) \to H_n(X, X-U_1;R) \oplus H_n(X,X-U_2;R) \to H_n(X, X - (U_1 \cap U_2);R),$$
where the left zero is because $\mathcal F_{n+1}$ is identically zero. This exactness is exactly the gluing property for $U_1$ and $U_2$, since the maps are induced by the inclusions of pairs, which are exactly the maps of $\mathcal F_n$.

Remarks. Just this is enough to show that if $X$ is compact, then $H_n(X;R)$ is isomorphic to global sections of the $R$-orientation cover $X_R = \{(x, \mu_x) \mid \mu_x \in H_n(X, X-x;R)\} \to X$. For this is true on a basis of balls, and $X$ can be covered by finitely many such balls, at which point finite gluing of $\mathcal{F}_n$ and gluing of sections may be applied.
