# Is this topological transformation group locally path connected?

A surface is an oriented connected sum of $$g\geq 0$$ tori, with $$b \geq 0$$ open disks removed, and $$n \geq 0$$ punctures in its interior.

Let Aut$$^+(S,\partial S)$$ denote the group (under composition) of orientation preserving homeomorphisms from $$S$$ onto itself which restrict to the identity map on the boundary $$\partial S$$. This is endowed with the compact open topology. Let Aut$$_0(S,\partial S)$$ denote the connected component of $$\mathrm{id}:S\to S$$.

In A Primer On Mapping Class Groups, the mapping class group $$\mathrm{Mod}(S)$$ of a surface $$S$$ is defined in the following two ways:

1. $$\mathrm{Mod}(S) = \pi_0(\mathrm{Aut}^+(S,\partial S), \mathrm{id})$$
2. $$\mathrm{Mod}(S) = \mathrm{Aut}^+(S,\partial S)\ /\ \mathrm{Aut}_0(S,\partial S)$$.

It is easy to show that the first definition corresponds to the set of boundary-fixing isotopy classes of maps in Aut$$^+(S,\partial S)$$, for which the group operation is simply $$[f][g] = [f\circ g]$$. However, I'm struggling to show that the second definition makes sense, let alone that it is equivalent to the first.

My problem is this: for the definitions to be equivalent, quotienting by $$\mathrm{Aut}_0(S,\partial S)$$ must correspond to quotienting by isotopy. Thus $$\mathrm{Aut}_0(S,\partial S)$$ must be the isotopy class of the identity. It is easy to see that this is just the path component of the identity. But why should I expect this to be the same as the connected component of the identity? I've tried to show that Aut$$^+(S,\partial S)$$ is locally path connected (which I believe is true, since the definition of a "surface" doesn't allow for much pathology). However I haven't been able to make any progress.

I forgot all about the fact that I asked this question here, but I answered it myself a while back.

Definitions:

A surface $$S$$ is an oriented connected sum of $$g \geq 0$$ tori with $$b \geq 0$$ open disks removed, and $$n \geq 0$$ punctures in its interior. Homeomorphism classes of surfaces are in bijective correspondence with $$\{(g,b,n): g,b,n \geq 0\}$$.

Given a surface $$S$$, the group (under composition) of orientation preserving homeomorphisms that fix the boundary $$\partial S$$ is denoted by Aut$$^+(S,\partial S)$$. This is endowed with the compact-open topology. The path component of Aut$$^+(S,\partial S)$$ containing id$$_S$$ is denoted Aut$$_0(S,\partial S)$$.

Theorem:

Given a surface $$S$$, the following are three equivalent definitions of its mapping class group, Mod$$(S)$$:

1. Mod$$(S)_1 := \pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$$.
2. $$\mathrm{Mod} (S)_2$$ is the group of boundary-fixing isotopy classes of maps in $$\mathrm{Aut}^+(S,\partial S)$$.
3. $$\mathrm{Mod}(S)_3 := \mathrm{Aut}^+(S,\partial S)/ \mathrm{Aut}_0(S,\partial S)$$.

We first show that $$1$$ and $$2$$ are naturally isomorphic as sets. Explicitly, $$\pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$$ is the collection of homotopy classes of continuous maps $$\sigma: \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$$ such that $$\sigma (0) = \mathrm{id}_S$$. Note that $$\boldsymbol 2$$ denotes the discrete space $$\{0,1\}$$. Suppose $$\sigma, \tau : \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$$ belong to the same homotopy class. Then there is a continuous map $$H: \boldsymbol 2 \times [0,1] \to \mathrm{Aut}^+(S,\partial S)$$ such that $$H(1,0) = \sigma(1)$$, $$H(1,1) = \tau(1)$$, and $$H(0,t) = \mathrm{id}_S$$ for each $$t \in [0,1]$$. Consider the map $$F: S \times [0,1] \to S$$ defined by $$F(s,t) := H(1,t)(s)$$ for all $$s \in S, t \in [0,1]$$. Then $$F(s,0) = \sigma(1)(s)$$ and $$F(s,1) = \tau(1)(s)$$ for each $$s \in S$$. Given any $$t \in [0,1]$$, $$F(-,t) \in \mathrm{Aut}^+(S,\partial S)$$. Therefore $$F$$ is a boundary-fixing isotopy from $$\sigma(1)$$ to $$\tau(1)$$. In summary, given any homotopy class $$[\sigma] \in \pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$$, evaluation at $$1$$ gives a boundary-fixing isotopy class of maps in $$\mathrm{Aut}^+(S,\partial S)$$.

Conversely, suppose $$f,g \in \mathrm{Aut}^+(S,\partial S)$$ are isotopic. One can similarly construct a homotopy between the maps $$\sigma,\tau: \boldsymbol 2 \to \mathrm{Aut}^+(S,\partial S)$$ defined by $$\sigma(0)=\tau(0)= \mathrm{id}_S$$, $$\sigma(1) = f$$, $$\tau(1) = g$$.

We now define the group structure on $$\mathrm{Mod} (S)_2$$. I claim that $$[f][g] = [f\circ g]$$ for each $$[f],[g]\in \mathrm{Mod}(S)_2$$ defines a group structure on $$\mathrm{Mod} (S)_2$$. Let $$f_0,f_1,g_0,g_1 \in \mathrm{Aut}^+(S,\partial S)$$ such that $$[f_0] = [f_1]$$ and $$[g_0] = [g_1]$$. Then there are boundary-fixing isotopies $$F^f, F^g: S \times [0,1] \to S$$ such that $$F^f(s,i) = f_i(s)$$ and $$F^g(s,i) = g_i(s)$$ for all $$s \in S$$. The map $$F: S \times [0,1] \to S$$ defined by $$F(s,t) = F^f(F^g(s,t),t)$$ for all $$s\in S, t\in [0,1]$$ is then a boundary-fixing isotopy from $$f_0 \circ g_0$$ to $$f_1 \circ g_1$$. Thus $$[f][g] = [f\circ g]$$ is well defined.

It is now routine to see that this binary operation is a group operation, where $$[\mathrm{id}_S] = e,\ \text{and}\ [f]^{-1} = [f^{-1}].$$ This also induces a group structure on $$\mathrm{Mod}(S)_1$$: For any $$[\sigma],[\tau] \in \pi_0(\mathrm{Aut}^+(S,\partial S), \mathrm{id})$$, $$[\sigma][\tau] = [\gamma],\ \text{where}\ \gamma(1) = \sigma(1) \circ \tau(1).$$

Finally, we determine how definitions $$1$$ and $$2$$ of $$\mathrm{Mod}(S)$$ correspond to $$\mathrm{Mod}(S)_3$$. $$\mathrm{Aut}^+(S,\partial S)$$ is naturally equipped with a group structure under composition. We first show that $$\mathrm{Aut}_0(S,\partial S)$$ is a normal subgroup of $$\mathrm{Aut}^+(S,\partial S)$$, and then that the quotient group $$\mathrm{Aut}^+(S,\partial S)/\mathrm{Aut}_0(S,\partial S)$$ is isomorphic to $$\mathrm{Mod}(S)_2$$.

Observe that there is a one-to-one correspondence between isotopies $$F: S \times [0,1] \to S$$ and paths $$\gamma : [0,1] \to \mathrm{Aut}^+(S,\partial S)$$. Given an isotopy $$F$$, simply define the path by $$t \mapsto (s \mapsto F(s,t)).$$ Thus $$\mathrm{Aut}_0(S,\partial S)$$ is the isotopy class of $$\mathrm{id}_S$$. From the above discussion about the group operation on $$\mathrm{Mod}(S)_2$$, it is now clear that $$\mathrm{Aut}_0(S,\partial S)$$ is a normal subgroup of $$\mathrm{Aut}^+(S,\partial S)$$. Thus the quotient is truly a group.

I now claim that $$\varphi: \mathrm{Mod}(S)_3 \to \mathrm{Mod}(S)_2$$ defined by $$f + [\mathrm{id}] \mapsto [f]$$ is a group isomorphism. Suppose $$f,g \in \mathrm{Aut}^+(S,\partial S)$$ such that $$f+[\mathrm{id}] = g+[\mathrm{id}]$$. Then there exists $$h \in \mathrm{Aut}_0(S,\partial S)$$ such that $$f = g \circ h$$. Then $$[f] = [g\circ h] = [g][h] = [g][\mathrm{id}] = [g]$$, so $$\varphi$$ is well defined. Clearly the map is surjective and injective. Finally, given any $$f+[\mathrm{id}], g+[\mathrm{id}]$$, $$\varphi(f+[\mathrm{id}])\varphi(g+[\mathrm{id}]) = [f][g] = [f\circ g] = \varphi (f\circ g + [\mathrm{id}]) = \varphi((f + [\mathrm{id}])(g + [\mathrm{id}])).$$ Therefore $$\varphi$$ is an isomorphism as required. This proves the equivalence of the three definitions for $$\mathrm{Mod}(S)$$.