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$(S)$:
- Mod$(S)_1 := \pi_0 (\mathrm{Aut}^+(S,\partial S), \mathrm{id}_S)$.
- $\mathrm{Mod} (S)_2$ is the group of boundary-fixing isotopy classes of maps in $\mathrm{Aut}^+(S,\partial S)$.
- $\mathrm{Mod}(S)_3 := \mathrm{Aut}^+(S,\partial S)/ \mathrm{Aut}_0(S,\partial S)$.
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 $\mathrm{Mod}(S)$.
Best Answer
According to Černavskiĭ, the local contractibility of $Homeo(M)$ for a noncompact topological manifold $M$ implies that $M$ could be the interior of a compact manifold with boundary. Of course this is true when $M=\mathbb{R}^n$. For a statement see p.8, Th.2, of his paper
A. Černavskiĭ, Local Contractibility of the Group of Homeomorphisms of a Manifold, 1969 Math. USSR Sb. 8 287.
Notes: If $M$ is a metrisable topological manifold, then $Homeo(M)$ in the compact-open topology is a metrisable topological group. If $M$ is compact, then $Homeo(M)$ is locally contractible. If $M$ is noncompact, then $Homeo(M)$ may fail to be locally contractible (cf. Kirby-Edwards). The implication is that $Homeo(M)$ for compact $M$ may be an ANR. This is known to be true in dimensions $\leq 2$ (cf. Yagasaki) however it is an open problem for dimensions $\geq3$.