$ (H, \alpha) $ is lie subgroup of $G$, $\exp(tY) \in \alpha(H)$, $\forall t \in \mathbb{R}$, How to prove $ Y \in \alpha_\ast (h)$? where $h$ is the lie algebra of $H$.
When I encounter this exc, the first thought in my mind is take the $t_0$ small enough, and the exponential map is homeomorphism in the neighborhood $V$ of the identity, therefore $$\exp(t_0Y) = \alpha(X) = \alpha(\exp(Y^{\prime})) = \exp(\alpha_\ast(Y^{\prime}))$$
hence, $t_0Y = \alpha_\ast(Y^{\prime}) \in \alpha_\ast(h)$.
But the $\exp(tY)$ may not be the uniparametric subgroups of $ \alpha(H) $, and $ \alpha(H) $ may not be lie group. I have no idea to work out this question. And I appreciate all your help.
The map $t\rightarrow \exp(tY)$ is a differentiable map defined on $\mathbb{R}$ and takes its values in $\alpha(H)$. This implies that its differential at $t$ is a map $T_t\mathbb{R}=\mathbb{R}\rightarrow \alpha(H)_{\exp(tY)}$ in particular if $t=0$ we have: ${d\over{dt}}_{t=0}\exp(tY)=Y$ is an element of $T_{Id}\alpha(H)=\alpha_*(h)$.