Let $G$ be Lie group with identity $e$ and dimension $n$. Let $G^0$ be the identity component of $G$, and let $\langle S \rangle$ be the subgroup generated by a subset $S$, which need not be a subgroup or submanifold, of $G$.
Consider the exponential map $\exp: T_eG \to G$. Let $U$ be an open neighbourhood of $e$ in $G$ ($U$ need not be homeomorphic to an open subset of $\mathbb R^n$). Then $T_eU$ is isomorphic to $T_eG$ under $i_{\{*,e\}}: T_eU \to T_eG$ where $i: U \to G$ is the inclusion map.
Questions:
Is it true that $\exp(i_{\{*,e\}}(T_eU)) \subseteq U$? (Equivalently, $\exp(T_eU) \subseteq U$, once you identify $T_eU$ with $i_{\{*,e\}}(T_eU)$.)
If no, then what if $U$ is homeomorphic to an open neighbourhood of $\mathbb R^n$?
If still no, then what if $U$ is homeomorphic to $\mathbb R^n$? (I think that if $U$ is homeomorphic to $\mathbb R^n$, then $U$ is diffeomorphic to $\mathbb R^n$, so there's no need to inquire further for the case that $U$ is diffeomorphic to $\mathbb R^n$)
All I know so far is that
If $U$ is connected (such as when $U$ is homeomorphic to $\mathbb R^n$), then $U \subseteq G^0$, by this and $\langle U \rangle = G^0$.
If $H$ is an open subgroup of $G$, then $H \supseteq G^0$
Image of $\exp$ is a subset of $G^0$.
Considering $T_eG$ as a manifold (diffeomorphic and isomorphic, possibly Lie group isomorphic, to $\mathbb R^n$), $i_{\{*,e\}}(T_eU)$ is an open subset of $T_eG$ that is diffeomorphic to (and I guess isomorphic and even Lie group isomorphic to) $\mathbb R^n$, even if $U$ is not homeomorphic to $\mathbb R^n$.
For every $V$ open in $T_eG$ (such as $V = i_{\{*,e\}}(T_eU)$, I think), $\langle \exp(V) \rangle = G^0$
Note: Please try not to use anything like $\exp$ is a local diffeomorphism at $Z_e$ because that's what I'm trying to prove (here: Differential of exponential map at identity)
Thanks in advance!