How to proof that a manifold covered by two charts satisfying $f_i(\mathbb{R}^m \setminus B^o) \subset f_j(B^o)$ for $i\not=j$ is simply connected? While studying for my comprehensive exam, I came across the following problem, which I am unable to solve.
Let $B$ denote the closed unit ball centered at the origin in Euclidean space $\mathbb{R}^m$, $\partial B$ denote the boundary sphere of $B$ and $B^o$ denote the interior of $B$.
Let $f_i: \mathbb{R}^m \to M$, $i=1,2$ denote smooth embeddings into a smooth $m$-dimensional manifold $M$ which satisfy
$$f_1(\mathbb{R}^m) \cup f_2(\mathbb{R}^m) = M$$
$$f_1(\mathbb{R}^m \setminus B^o) \subset f_2(B^o) $$
$$f_2(\mathbb{R}^m \setminus B^o) \subset f_1(B^o) $$
Suppose that $m \geq 2$. Show that $M$ is a simply connected closed manifold.

Here is what I have been able to do so far.
Proof that M is closed:
Let $\{x_i\}$ be a sequence in $M$, then we can subtract a subseqeunce $\{x_{i_j}\}$ which is contained entirely within $f_1(\mathbb{R}^m)$ or entirely within $f_2(\mathbb{R}^m)$. Suppose without loss of generality that $\{x_{i_j}\}$ is contained entirely within $f_1(\mathbb{R}^m)$, then we can subtract a subsubsequence $\{x_{i_{j_k}}\}$ which is contained entirely within $f_1(B)$ or entirely within $f_1(\mathbb{R}^m \setminus B^o)$. Suppose first that $\{x_{i_{j_k}}\}$ is contained entirely within $f_1(B)$ then by compactness of $f_1(B)$ there is subsubsubsequence $\{x_{i_{j_{k_l}}}\}$ which converges in $f_1(B)$. If instead the subsubsequence $\{x_{i_{j_k}}\}$ is contained entirely within $f_1(\mathbb{R}^m \setminus B^o)$, then by assumption the subsubsequence is contained within $f_2(B)$, which is compact, so there is a subsubsubsequence $\{x_{i_{j_{k_l}}}\}$ which converges. Hence $M$ is compact. Since $f_1$, and $f_2$ are charts without boundary that covers $M$, then $M$ is a manifold without boundary, so $M$ is closed.
I am unable to show that $M$ is simply connected. It is connected, as it is the union of two connected sets with non-empty intersection. If $f_1(\mathbb{R}^m) \cap f_2(\mathbb{R}^m)$ is connected then by the Seifert-Van Kampen Theorem $M$ would be simply connected. My idea has been to try and come up with suitable open subsets $U, V$ and then use the Mayer-Vietoris sequence to show that the 0th de Rham Cohomology group of $f_1(\mathbb{R}^m) \cap f_2(\mathbb{R}^m)$ has dimension 1. None of my candidates for $U$ and $V$ has worked so far.
The result fails when $m=1$ since then $M=S^1$. I am not sure how to invoke the $m \geq 2$ hypothesis, if the problem is to be solved without a Mayer-Vietoris argument.
The problem should be solvable using techniques from Topology (By Munkres) and Introduction to Smooth Manifolds (By Lee). This means no Generalized Jordan Curve Theorem, Reduced Integral Homology or Alexander Duality. Any help/solution would be much appreciated.
 A: Here’s a shorter argument about why $M$ is closed. It’s enough to show that $M$ is a compact topological space. But by your hypothesis $M=f_1(B) \cup f_2(B)$ where $B$ is the closed unit ball (compact) and the $f_i$ are continuous.
About simple connectedness: note that $M=f_1(B^\circ) \cup f_2(B^\circ)$ is an open cover with two simply connected (contractible, really!) open subsets. By van Kampen, it’s enough to show that $f_1(B^\circ) \cap f_2(B^\circ)$ is connected.
Now, by continuity of the $f_i$ and openness, there exists $r <1$ such that $f_i((r,r^{-1})\partial B) \subset f_{3-i}(rB^\circ)$.
It is enough to show that $U_0=B^\circ \cap f_2^{-1}(f_1(B^\circ))$ is connected. Note that $U_0 \supset (r,1)\partial B:=U_1$ (let $y$ be some point in $(r,1)\partial B$. Now, let $x \in U_0$. Then $f_1^{-1}(f_2(x)) \in B^\circ$ so there is a path $p:[0,1] \rightarrow M$ from $f_2(x)$ to $f_2(y)$ not meeting $f_1(\partial B)$.
Let $I=\{t \in [0,1],\, p([0,t]) \subset f_2(B^\circ)\}$. $I$ is the connected component of $0$ in $p^{-1}(f_2(B^\circ))$ so it is an open interval (ie open in $[0,1]$) containing zero. Now, $I$ is contained in the compact $p^{-1}(f_2(B))$, so that if $s$ is the supremum of $I$, there is $0<t\leq s$ such that $t \in I$ and $\|f_2(p(t))\| > r$.
(If $s=1$, $t=1$ works).
Then $f_2^{-1} \circ p$ is defined on $[0,t]$ and has values in $B^\circ \backslash f_2^{-1}(f_1(\partial B))$, so, since $x \in f_2^{-1}(f_1(B^\circ))$, $p$ has values in $f_1(B^\circ)$ as well, so that $x$ is connected to $U_1$ in $U_0$.
So all of $U_0$ is connected to $U_1$, which is connected, so $U_0$ is connected. QED.
