A connected manifold can be covered by finite open subsets

Question

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

Answer 1

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:

1. 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.

2. $$\bigcup_{i=1}^{n+1} \bigcup_{j\in J_i} int(B_{ij})=M.$$

3. 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.