# Dehn-Nielson-Baer and Basepoints

Let $S$ be an orientable compact closed surface. Fix a base point $x \in S$ . Then given a homeomorphism $f : S \to S$, we have an induced isomorphism $f_\ast : \pi_1(S, x) \to \pi_1(S,f(x))$. By choosing an arc $\alpha$ from $x$ to $f(x)$, we obtain an automorphism $\alpha_\sharp \circ f_\ast : \pi_1(S, x) \to \pi_1(S, x)$. Different choices of this arc $\alpha$ result in different automorphisms of $\pi_1(S,x)$ however if a different arc $\alpha'$ is chosen then the two resulting automorphisms differ by conjugation by the loop given by $\alpha$ and $\alpha'$. Therefore, we have a homomorphism $$\text{Homeo}(S) \to \text{Out}(\pi_1(S,x))$$ which the Dehn-Nielson-Baer theorem says is an isomorphism.

My question is this: given an automorphism $f : \pi_1(S,x) \to \pi_1(S,x)$ does there exist a base point preserving homeomorphism $F : (S,x) \to (S,x)$ where $F$ induces $f$ ??

To prove it, from the Dehn-Nielsen-Baer theorem there exists a homeomorphism $F' : S \to S$ and a path $\alpha$ from $x$ to $f(x)$ such that the outer automorphism classes of $\alpha_\# \circ F'_*$ and of $f$ are equal. Thus, there exists a closed path $\beta$ based at $x$ such that $$f = \beta_\# \circ \alpha_\# \circ F'_* = (\beta * \alpha)_\# \circ F'_* = \gamma_\# \circ F'_*$$ where $\gamma = \beta * \alpha$. Let $P_{\bar\gamma}$ be the "point pushing" homeomorphism associated to the path $\bar\gamma$: starting from the identity, isotope the surface by slowing pushing the point $f(x)$ forward along the path $\bar\gamma$ --- equivalently, backwards along $\gamma$. Then $F = P_{\bar\gamma} \circ F'$ fixes $x$ and induces $f$.