Krdytkyu

A surface is an oriented connected sum of $ggeq 0$ 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 $mathrm{id}:Sto S$.

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 $[f][g] = [fcirc g]$. 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
$$[f][g] = [fcirc g]$$
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 $sin S, tin [0,1]$ is then a boundary-fixing isotopy from $f_0 circ g_0$ to $f_1 circ g_1$. Thus $[f][g] = [fcirc g]$ 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 $[f] = [gcirc h] = [g][h] = [g][mathrm{id}] = [g]$, so $varphi$ is well defined. Clearly the map is surjective and injective. Finally, given any $f+[mathrm{id}], g+[mathrm{id}]$,
$$varphi(f+[mathrm{id}])varphi(g+[mathrm{id}]) = [f][g] = [fcirc g] = varphi (fcirc g + [mathrm{id}]) = varphi((f + [mathrm{id}])(g + [mathrm{id}])).$$
Therefore $varphi$ is an isomorphism as required. This proves the equivalence of the three definitions for $mathrm{Mod}(S)$.