Why is the restricted holonomy the identity component of the holonomy group? Let $M$ be a connected smooth paracompact manifold, $E$ a vector bundle over $M$ with fibre $\mathbb R^k$, and $\nabla$ a connection on $E$. It is known that Hol$^0(\nabla)$ is a connected Lie subgroup of $GL(k,\mathbb R)$. How can we show Hol$^0(\nabla)$ is an identity component of Hol$(\nabla)$? 
It seems to me that there are two ways to understand this. The first way is to regard Hol$^0(\nabla)$ and Hol$(\nabla)$ as topological subspaces of $GL(k,\mathbb R)$. Another way is to make Hol$(\nabla)$ a Lie subgroup of $GL(k,\mathbb R)$ by left translating the differential structure of Hol$^0(\nabla)$. But to prove Hol$(\nabla)$ is a Lie group, one has to prove for any $a\in$ Hol$(\nabla)$, the mapping from Hol$^0(\nabla)$ to Hol$^0(\nabla)$ defined by $x \rightarrow axa^{-1}$ is differentiable. I am stuck here.
 A: Let $L$ be a loop which is contractible to the constant loop $P$ (at base point $p$). This means that there is a homotopy $L_s, s\in[0,1]$ such that $L_0=P$ and $L_1=L$, continuous with respect to $s$. 
Let $hol(L)\in GL(k)$ denote the holonomy around the loop $L$.
We have a very good mapping $s\mapsto hol(L_s)$, from $[0,1]$ to $GL(k)$, which is continuous! In other words, a continuous curve in $GL(k)$, or more specifically, in $Hol_0(\nabla)$. 
Since $hol(L_0)=1$, the image contains the identity. And a continuous path connects it to $hol(L)$.
If we construct this for all the contractible loops $L$, by definition we span exactly the group $Hol_0(\nabla)$. Which is therefore path-connected, and it contains the identity. 
---this may already answer your question.
Now some non-contractible loops, of course, can still be mapped to the restricted holonomy group. But I want to prove that, at least on manifolds which are "nice enough", those loops that are not mapped into $Hol_0(\nabla)$ are mapped to a place in $GL(k)$ which is not path-connected to the identity. In other words, the rest of $Hol(\nabla)$ is not path connected to $Hol_0(\nabla)$. (Think of the points in $O(2)$ which are not in $SO(2)$.)
In particular, consider a non-contractible loop $L$. Suppose that there exists a continuous curve $\gamma:[0,1]\to Hol(\nabla)$ such that $\gamma(0)\in Hol_0(\nabla)$ and $\gamma(1)=L$. 
This means that there are loops $L_s:[0,1]\to Hol(\nabla)$, for which $hol(L_s)=\gamma(s)$, and that there is a contractible loop $M$ such that $hol(M)=hol(L_0)=\gamma(0)$. 
Now, in the best case, the loops $L_s$ are homotopic (and $s\mapsto L_s$ is a homotopy). Consider the loop $L_sL_0^{-1}$, given by the concatenation of $L_s$ with the inverse of $L_0$ (inverse meaning "walked backwards"). It is of course a contractible loop, so $hol(L_sL_0^{-1})\in Hol_0(\nabla)$. But:
$$
hol(L_sL_0^{-1}) = hol(L_s)hol(L_0)^{-1},
$$
and since $hol(L_0)\in Hol_0(\nabla)$, we conclude that $hol(L_s)\in Hol_0(\nabla)$, and in particular $hol(L)\in Hol_0(\nabla)$.
If the loops $L_s$ are not homotopic, the second-best case is that we can split the counter-image of our curve $\gamma$ into two different homotopies, say $\{L_s, s\le 1/2\}$ and $\{K_s, s\ge 1/2\}$, with $L_{1/2}\ne K_{1/2}$, but $hol(L_{1/2})=hol(K_{1/2})=\gamma(1/2)$.
In this case you can proceed like in the previous case, but twice: first you use $L_sL_0^{-1}$ to show, exactly like before, that $hol(L_s)\in Hol_0(\nabla)$. You conclude that $hol(L_{1/2})=hol(K_{1/2})\in Hol_0(\nabla)$. Then you use $K_sK_{1/2}^{-1}$ to show that $hol(K_s)\in Hol_0(\nabla)$. You conclude again that $K_1=L$ has holonomy in $Hol_0(\nabla)$.
You can iterate this process three times, four, etcetera, as needed. So this works if you can get from $L_0$ to $L$ in a finite number of homotopies. 
The curve $\gamma$, therefore, is split into a finite number of segments each of which is the image of a homotopy of loops.
Since loops are compact (these are the loops you work with in holonomy), it is almost guaranteed that you only need a finite number of steps. This probably doesn't work if there are too many "holes", meaning, if the fundamental group of our base manifold is not finitely generated.
We have proved that (at least on "usual" manifolds), if an element of $Hol(\nabla)$ admits a path connecting it to an element $Hol_0(\nabla)$, then it lies in $Hol_0(\nabla)$.
