In Vick's $\textit{An Introduction to Algebraic Topology}$,
6.5 theorem states that “if M is a connected, noncompact n-manifold without boundary, then $H_n(M) = 0$".
Consequently, I have a question:
If M is a connected n-manifold without boundary, can M be covered by finite open subsets $\{U_1, \dots , U_k\}$ such that every $U_i$ is homeomorphic to $ \mathbb{R}^n$?
Even though I have tried many examples, but I didn't construct a counterexample. I don't know whether it is right.
Here are some related results:
(1)https://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover
(2)Surface where number of coordinate charts in atlas has to be infinite
Best Answer
Good question. The answer is positive.
First of all, as in my answer here, there exist $n+1$ families ${\mathcal F}_1,..., {\mathcal F}_{n+1}$ of subsets in $M$ such that:
Each ${\mathcal F}_i$ is a union of closed (and tame) pairwise disjoint $n$-balls $B_{ij}, j\in J_i$, where each $J_i$ is (at most) countable.
$$\bigcup_{i=1}^{n+1} \bigcup_{j\in J_i} int(B_{ij})=M.$$
Each compact $K$ in $M$ intersects only finitely many closed balls in the above collections.
I will show that each $$ V_i= \bigcup_{j\in J_i} int(B_{ij}) $$ is contained in an (open) subset $U_i$ of $M$ homeomorphic to $R^n$. This will do the job. To construct $U_i$, I will treat each $B_{ij}$ as a $0$-handle in $M$. I will identify each $J_i$ with an interval in ${\mathbb N}$ (finite or infinite). Next, for each pair of consecutive indices $j, j+1\in J_i$, connect $B_{ij}, B_{ij+1}$ by a 1-handle $H_{ij}$ in $M$ so that distinct 1-handles are pairwise disjoint and intersect only the balls $B_{ij}, B_{ij+1}$ in ${\mathcal F}_i$ and only along disks in their boundaries. A 1-handle is a thickened (tame, simple) arc $a_{ij}$ in $M$ connecting boundary spheres of $B_{ij}, B_{ij+1}$. Making these arcs $a_{ij}$ pairwise disjoint is easier if $M$ has dimension $\ge 3$ (first take any locally finite collection of tame arcs and then away any accidental intersection). Constructing these arcs in the case of surfaces is not hard but tedious, I can explain how to do so if you like. (Connectivity of $M$ is used to ensure the existence of an arc connecting $B_{ij}, B_{ij+1}$.)
Informally, the union of the 0-handles $B_{ij}$ and $1$-handles $H_{ij}$ is a "chain of closed balls" (finite or infinite). Arguing inductively, one sees that each finite chain $$ W_i:=B_{i1}\cup H_{i1}\cup B_{i2} \cup ... \cup H_{i,k-1}\cup B_{ik} $$ is homeomorphic to the closed $n$-ball. Similarly, each infinite chain $$ W_i:=B_{i1}\cup H_{i1}\cup B_{i2} \cup ... \cup H_{i,k-1}\cup B_{ik} \cup ... $$ is homeomorphic to the closed $n$-dimensional half-space. The interior $U_i$ of each chain is homeomorphic to ${\mathbb R}^{n+1}$.
Thus, $$ M= U_1\cup ...\cup U_{n+1}. $$
A similar argument works in PL and smooth categories.