For the sake of this problem a manifold of dimension \(n\) will be defined as a topological space in which each point has a neighborhood that is homeomorphic to \(\mathbb R^n\). If \(M\) is a connected manifold of dimension at least \(3\) and \(q\in M\), show that \(\pi_1(M-\{q\})\) is isomorphic to \(\pi_1(M)\).
Solution
Let \[ V=M\setminus\{q\}. \]
1. Every singleton in \(M\) is closed; in particular, \(V\) is open in \(M\).
Fix distinct points \(x,y\in M\). Choose a Euclidean neighborhood \(N\) of \(y\) and an open set \(O\subseteq M\) with \[ y\in O\subseteq N. \] If \(x\notin N\), then \(O\) is already an open neighborhood of \(y\) avoiding \(x\).
If \(x\in N\), the space \(N\) is homeomorphic to \(\mathbb R^n\) and hence is \(T_1\). Thus there is a set \(W\) open in the subspace \(N\) such that \[ y\in W \qquad\text{and}\qquad x\notin W. \] Write \[ W=N\cap G \] for some open \(G\subseteq M\). Then \[ O\cap G \] is an open neighborhood of \(y\) in \(M\) that avoids \(x\).
Thus, for every \(y\ne x\), there is an open neighborhood of \(y\) disjoint from \(\{x\}\). Hence \[ M\setminus\{x\} \] is open, so \(\{x\}\) is closed. Taking \(x=q\) proves that \(V\) is open.
2. We may choose an open neighborhood \(U\) of \(q\) homeomorphic to \(\mathbb R^n\), with \(q\) corresponding to \(0\).
Let \(N\) be a neighborhood of \(q\) homeomorphic by \[ h:N\longrightarrow\mathbb R^n \] to Euclidean space. Since \(N\) is a neighborhood, choose open \(O\subseteq M\) with \[ q\in O\subseteq N. \] The set \(h(O)\) is open in \(\mathbb R^n\) relative to the chart, so it contains an open Euclidean ball \(B\) about \(h(q)\). After translating coordinates, take \(h(q)=0\). Then \[ U=h^{-1}(B) \] is open in \(M\): because \(h^{-1}(B)\) is open in the subspace \(N\), write it as \(N\cap G\) for some open \(G\subseteq M\); since it lies in \(O\subseteq N\), it equals \(O\cap G\). Finally, composing \(h|_U\) with a homeomorphism \(B\cong\mathbb R^n\) gives the desired open chart \[ U\cong\mathbb R^n \] sending \(q\) to \(0\).
3. The punctured chart\[ U\setminus\{q\} \]is path-connected.
By 2, \[ U\setminus\{q\} \cong \mathbb R^n\setminus\{0\}. \] The radial map \[ x\longmapsto\frac{x}{\|x\|} \] is the endpoint of the deformation retraction \[ H(x,t) =\left((1-t)+\frac{t}{\|x\|}\right)x \] from \(\mathbb R^n\setminus\{0\}\) onto \(S^{n-1}\). Since \(n\ge3\), the sphere \(S^{n-1}\) has dimension at least \(2\) and is path-connected. Therefore \(U\setminus\{q\}\) is path-connected.
4. The punctured manifold \(V=M\setminus\{q\}\) is path-connected.
Every point of \(M\) has a Euclidean neighborhood, so \(M\) is locally path-connected. The open subspace \(V\) is therefore locally path-connected as well. Hence every path component of \(V\) is open in \(V\), and by 1 is consequently open in \(M\).
By 3, the set \[ U\setminus\{q\} \] is path-connected, so it lies in a single path component \(C_0\) of \(V\). Suppose there were another path component. Let \(W\) be the union of all path components of \(V\) other than \(C_0\). Then \(W\) is a nonempty open subset of \(M\). Moreover, \[ C_0\cup\{q\}=C_0\cup U, \] because \(U\setminus\{q\}\subseteq C_0\). The right-hand side is open in \(M\), and it is the complement of \(W\). Thus \[ M=W\sqcup(C_0\cup\{q\}) \] would be a separation of \(M\), contradicting the connectedness of \(M\). Therefore \(W\) is empty and \(V=C_0\) is path-connected.
5. For \(n\ge3\), the punctured chart \(U\setminus\{q\}\) is simply connected.
By 3 it deformation retracts onto \[ S^{n-1}. \] It therefore suffices to show that \(S^m\) is simply connected for \(m\ge2\).
Cover \(S^m\) by two open sets \(A,B\), each obtained by enlarging an open hemisphere slightly past the equator. Both \(A\) and \(B\) are contractible. Their intersection deformation retracts onto the equatorial sphere \[ S^{m-1}, \] which is path-connected because \(m-1\ge1\). The Seifert–van Kampen theorem gives \[ \pi_1(S^m) \cong \pi_1(A)*_{\pi_1(A\cap B)}\pi_1(B). \] Since \[ \pi_1(A)=\pi_1(B)=0, \] this amalgamated product is the trivial group. Thus \(S^m\) is simply connected for \(m\ge2\), and hence so is \(U\setminus\{q\}\).
6. The inclusion\[ i:V=M\setminus\{q\}\hookrightarrow M \]induces an isomorphism on fundamental groups.
Choose a basepoint \[ x_0\in U\setminus\{q\}. \] By <1>1–<1>2, both \(U\) and \(V\) are open, and \[ M=U\cup V. \] By <1>3–<1>4, the sets \(U\), \(V\), and \[ U\cap V=U\setminus\{q\} \] are path-connected. Furthermore, \[ \pi_1(U,x_0)=0 \] because \(U\cong\mathbb R^n\), and by 5, \[ \pi_1(U\cap V,x_0)=0. \]
The Seifert–van Kampen theorem identifies \(\pi_1(M,x_0)\) with the pushout \[ \pi_1(U,x_0) *_{\pi_1(U\cap V,x_0)} \pi_1(V,x_0). \] Both groups involving \(U\) are trivial, so this pushout is canonically \[ \pi_1(V,x_0). \] Under this identification, the canonical map from \(\pi_1(V,x_0)\) to the pushout is exactly the homomorphism induced by the inclusion \(i\). Therefore \[ i_*:\pi_1(M\setminus\{q\},x_0) \xrightarrow{\ \cong\ } \pi_1(M,x_0). \] Hence \[ \boxed{\pi_1(M\setminus\{q\})\cong\pi_1(M).} \]