Let $G$ be a reductive complex connected Lie group and $\mathfrak{g}$ be its Lie algebra. Let us fix a set of simple roots $\Pi$, the set of positive roots $\Phi_+$ and a nondegenerate invariant symmetric bilinear form on $\mathfrak{g}$. Let $\mathfrak{h}$ be the Cartan and $\mathfrak{n}_{\pm}$ be the nilpotent subalgebras. Choose root vectors $e_{\alpha}$ such that $\langle e_{\alpha}, e_{-\alpha}\rangle = 1$. Define a map $t:\mathfrak{g}\rightarrow\mathfrak{g}$, $t|_{\mathfrak{h}} = \mathrm{id}|_{\mathfrak{h}}$, $t(e_{\alpha}) = e_{-\alpha}$, $t(e_{-\alpha}) = e_{\alpha}$. Then $t$ is an anti-homomorphism and an involution of $\mathfrak{g}$.
The question is: can $t$ be integrated to $G$ even without assuming simply-connectedness? Could there be any issue?
Upd.: Well, the map $t$ can be definitely integrated to a Zariski open subset of $G$. Let $\mathcal{N}_+$ and $\mathcal{N}_-$ be the unipotent radicals of opposite Borels along with the Cartan subgroup $\mathcal{H}$. On an open dense subset of $G$, any element $g \in G$ has a unique decomposition $g = g_+ g_0 g_-$, $g_\pm \in \mathcal{N}_\pm$, $g_0 \in \mathcal{H}$. Since $t$ can be integrated to the separate pieces $\mathcal{N}_\pm$ and $\mathcal{H}$, we can simply set $t(g_+g_0g_-) = t(g_-)t(g_0)t(g_+)$. Now the question boils down to the following: can $t$ be extended to the whole $G$?