The following is problem 10-6 of John M. Lees' Introduction to Topological manifolds:
Suppose $M$ is a connected manifold of dimension $n \geq 3$, and $p \in M$. Show that the inclusion $M\backslash \{p\} \hookrightarrow M$ induces an isomorphism $\pi_1(M\backslash \{p \}) \cong \pi_1(M)$
I can see that these fundamental groups are isomorphic by the van Kampen Theorem. Specifically, I can take a neighborhood $U_1$ of $p$ that is homeomorphic to an open ball in $\mathbb{R}^n$ and take $U_2$ to be $M\backslash \{p \}$. Both are open and both are path connected (the later is because $M$ is connected). Moreoever, the intersection of these two sets is a ball in $\mathbb{R}^n$ with a point removed and so is simply connected (it is here where we use that $n \geq 3$, as this is not true if $n = 1,2$). Thus, by van Kampen Theorem $\pi_1(M) = \pi_1(M\backslash \{p\})*\pi_1(U_1) $. But, $U_1$ is simply connected so $\pi_1(M) = \pi_1(M\backslash \{p \})$.
What I don't know is why the inclusion is an isomorphism. You can likely get surjectivity by quoting Lemma 7.19 which states that if $M$ is a manifold of dimension $\geq 2$ and $f$ is a path in $M$ from $p_1$ to $p_2$ and $q$ is any point in $M$ other than $p_1$ or $p_2$, then $f$ is path homotopic to a path that does not pass through $q$.
However, I do not know why the induced map should be injective. Any help or comments would be greatly appreciated.
Best Answer
The map $\pi_1(M\setminus p) \to \pi_1(M)$ from the van Kampen theorem is already induced by inclusion.
One way of seeing this (from a category theory perspective) is by noting that the van Kampen theorem actually concerns the pushout
$$\require{AMScd} \begin{CD} \pi_1(U_1\cap U_2) @>{\iota_1}>> \pi_1(U_2)\\ @V\iota_2VV @V\rho_2 VV \\ \pi_1(U_1) @>\rho_1>> P @>h>> \pi_1(U_1 \cup U_2) \end{CD}$$
where $\iota_1$ and $\iota_2$ are induced by inclusion, and $\rho_1$ and $\rho_2$ are given by the pushout. $P$ is typically denoted $\pi_1(U_1) \ast_{\pi_1(U_1\cap U_2)} \pi_1(U_2)$. The homomorphisms $j_1\colon\pi_1(U_1) \rightarrow \pi_1(U_1 \cup U_2)$ and $j_2\colon\pi_1(U_2) \rightarrow \pi_1(U_1 \cup U_2)$ induced by inclusion agree when precomposed with $\iota_1$ and $\iota_2$, so the universal property of the pushout gives us a natural homomorphish $h\colon P \to \pi_1(U_1\cup U_2)$ such that $j_1 = h\circ \rho_1$ and $j_2 = h \circ \rho_2$. The statement of van Kampen is then that (under the appropriate assumptions) $h$ is an isomorphism.
In your particular case, $U_1 \simeq *$, $U_2 = M\setminus p$, and $U_1 \cap U_2 \simeq S^{n-1}$, which is simply-connected when $n\geq 3$. In this case the pushout diagram has a single non-zero morphism $\rho\colon\pi_1(M\setminus p) \to P$, and the universal property reduces to the following statement: given any morphism $f\colon \pi_1(M\setminus p) \to G$ there is a unique morphism $h\colon P \to G$ such that $f = h\circ \rho$. You can use this simplified universal property to give an inverse for $\rho$ and show it is an isomorphism (exercise in universal properties). The homomorphism $i\colon \pi_1(M\setminus p) \to \pi_1(M)$ induced by inclusion gives us (by the universal property of the pushout) a homomorphism $h\colon P \to \pi_1(M)$ with the property that $i = h\circ \rho$, and by van Kampen theorem $h$ is an isomorphism. Therefore the map $\pi_1(M\setminus p) \to \pi_1(M)$ induced by inclusion factors as two isomorphisms.