Existence of a homeomorphism sending $p$ to $p'$, sending a generator of $\pi_1(U \setminus \{p\})$ to a generator of $\pi_1(U' \setminus \{p'\})$

Your approach has the right idea.

For $a \in \mathbb R^2$ and $r > 0$ let

• $B_r(a) = \{ x \in \mathbb R^2 \mid \lVert x -a \rVert < r\}$ = open disk with center $a$ and radius $r$
• $D_r(a) = \{ x \in \mathbb R^2 \mid \lVert x - a \rVert \le r\}$ = open disk with center $a$ and radius $r$.

A coordinate ball on $M$ is an open $U \subset M$ which is homeomorphic to some $B_r(a)$. Given a coordinate ball $U$ and $p \in U$, we can find a homeomorphism $h : U \to B_1(0)$ such that $h(p) = 0$. Call this a normal chart for $(U,p)$.

For the given coordinate balls $U, U'$ around $p, p'$ take normal charts $h : U \to B_1(0)$ and $h' : U \to B_1(0)$. Since $U \setminus \{p\}$ and $U' \setminus \{p'\}$ are homeomorphic to $B_1(0) \setminus \{0\}$, their fundamental groups are isomorphic to $\mathbb Z$.

For $r$ with $0 < r < 1$ let $D_{h,r} = h^{-1}(D_r(0))$ and $D_{h',r} = (h')^{-1}(D_r(0))$. These are strong deformation retracts of $U$ and $U'$.

We know that there exists a homeomorphism $F : M \to M'$ taking $p$ to $p'$. There exist open neighborhoods $V$ of $p$ in $M$ and $V'$ of $p'$ in $M$ such that $V \subset U, V' \subset U'$ and $F(V) = V'$. There exists $r'$ such that $D_{h',r'} \subset V'$ and $r$ such that $D_{h,r} \subset F^{-1}(D_{h',r'})$ (note that $F^{-1}(D_{h',r'})$ contains an open neighborhood of $p$).

The loop $u : [0,1] \to B_1(0) \setminus \{0\} , u(t) = r(\cos 2\pi t, \sin 2\pi t)$, gives a generator $[u]$ of $\pi_1(B_1(0) \setminus \{0\}, p_0)$ with $p_0 = (r,0)$, thus $f = h^{-1} \circ u$ gives a generator $[f]$ of $\pi_1(U \setminus \{p\}, x_0)$ with $x_0 = h^{-1}(p_0)$. We have $F(f([0,1]]) \subset F(D_{h,r}) \subset F(F^{-1}(D_{h',r'})) \subset F^{-1}(F(V')) =$ $V' \subset U'$. Since $p \notin f([0,1])$, we see that $p' = F(p) \notin F(f([0,1]])$, thus $g = F \circ f : [0,1] \to U' \setminus \{p'\}$ is well-defined. We shall prove that $[g]$ is a generator of $\pi_1(U' \setminus \{p'\}, y_0)$ with $y_0 = F(x_0)$.

The image of $f$ is contained in $D_{h,r} \setminus \{p\}$ and the image of $g$ in $D_{h',r'} \setminus \{p'\}$. Let $\rho : U \setminus \{p\} \to D_{h,r} \setminus \{p\}$ and $\rho' : U' \setminus \{p'\} \to D_{h',r'} \setminus \{p'\}$ be retractions.

Then $\phi : (U \setminus \{p\}, x_0) \xrightarrow{\rho} (D_{h,r} \setminus \{p\}, x_0) \xrightarrow{F} (U' \setminus \{p'\}, y_0)$ and $\psi : (U' \setminus \{p'\}, y_0) \xrightarrow{\rho'} (D_{h',r'} \setminus \{p'\}, y_0) \xrightarrow{F^{-1}} (U \setminus \{p\}, x_0)$ are basepoint preserving maps. Consider the induced homomorphisms

$$\phi_* :\pi_1(U \setminus \{p\}, x_0) \to \pi_1(U' \setminus \{p'\}, y_0) ,$$ $$\psi_* :\pi_1(U' \setminus \{p'\}, y_0) \to \pi_1(U \setminus \{p\}, x_0) .$$

We have $\phi_*([f]) = [\phi \circ f] = [F \circ \rho \circ f]$. Since $\rho = id$ on $D_{h,r} \setminus \{p\}$, we see that $\phi_*([f]) = [g]$. Similarly $\psi_*([g]) = [\psi \circ g] = [F^{-1} \circ \rho' \circ g] = [F^{-1} \circ g]$. But$ F^{-1} \circ g = f$, thus $\psi_*([g]) = [f]$.

Let $\gamma$ be a generator of the infinite cyclic group $\pi_1(U' \setminus \{p'\}, y_0)$. Then $[g] = n\gamma$ with soem $n \in \mathbb Z$.

We claim that $n = \pm 1$ which shows that $[g]$ is a generator of $\pi_1(U' \setminus \{p'\}, y_0)$.

We have $\psi_*(\gamma) = m[f]$ with some $m \in \mathbb Z$. Therefore $$[f] = \psi_*([g]) = \psi_*(n\gamma) = n \psi_*(\gamma) = nm[f] .$$ This implies $nm = 1$, thus $n = m = 1$ or $n = m = -1$.