Fiber-connected Lie subgroupoid of a Lie groupoid $\mathcal{G}\rightrightarrows M$ is open Let $\mathcal{G}\rightrightarrows M$, which has possible disconnected $\mathbf{t},\mathbf{s}$-fibers and disconnected set of units. We denote by $\mathcal{G}^{(0)}$ the union of all connected components of the fibers $\mathbf{t}^{-1}(x)$ containing the unit $1_x$. Then $\mathcal{G}^{(0)}$ is supposed to be an open Lie subgroupoid of $\mathcal{G}$, so they have the same Lie algebroid asscoiated to them. I wanted to show that this set is indeed open, so I wanted to construct an open neighbourhood $U$ around every $g\in \mathcal{G}^{(0)}$ such that any $h\in U$ can be connected to $1_{\mathbf{t}(h)}$ by a parth completely lying in $\mathbf{t}^{-1}(\mathbf{t}(h))$, but I don't really know what to work with.
 A: Here is my attempt. My approach does not use paths, so I don't know how much use of it is to you.
To show that $\mathcal{G}^{(0)}$ is open, I want to construct an open around each $g\in \mathcal{G}^{(0)}$. I think it is sufficient to construct an open $U$ around the identity section $\textbf{u}(M)\subseteq \mathcal{G}$ such that it is contained in $\mathcal{G}^{(0)}$ and then move around this open to any point in $\mathcal{G}^{(0)}$ using left multiplication.
Since $\textbf{s}:\mathcal{G}\rightarrow M$ is a submersion we can apply the local normal form theorem (see the wiki page on submersions). This says that for each $g\in\mathcal{G}$ there is an open neighborhood $U_g$ of $g$ together with an open neighborhood $V_{\textbf{s}(g)}$ of $\textbf{s}(g)$ such that $\textbf{s}(U_g)=V_{\textbf{s}(g)}$. Applying this to the identity $g=1_x\in\mathcal{G}^{(0)}$ we obtain open neighborhoods $U_x$ around all the units $1_x\in\mathcal{G}^{(0)}$. (I guess these $U_x$ need not be connected, but maybe they always intersect the s-fibers $s^{-1}(x)$ in a connected set, so $U_x\subseteq\mathcal{G}^0$.) Our neighborhood around the identity section $\textbf{u}(M)\subseteq\mathcal{G}$ is then $$U=\bigcup_{x\in M}U_x.$$
(If you have a more formal argument for why we can move $U$ around $\mathcal{G}^{(0)}$ please post this, because I would like to know. The same for why $U_x\subseteq\mathcal{G}^0$.)
