# Image of restriction of exponential map $\exp(T_eU) \subseteq U$?

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:

1. Is it true that $$\exp(i_{\{*,e\}}(T_eU)) \subseteq U$$? (Equivalently, $$\exp(T_eU) \subseteq U$$, once you identify $$T_eU$$ with $$i_{\{*,e\}}(T_eU)$$.)

2. If no, then what if $$U$$ is homeomorphic to an open neighbourhood of $$\mathbb R^n$$?

3. 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

1. If $$U$$ is connected (such as when $$U$$ is homeomorphic to $$\mathbb R^n$$), then $$U \subseteq G^0$$, by this and $$\langle U \rangle = G^0$$.

2. If $$H$$ is an open subgroup of $$G$$, then $$H \supseteq G^0$$

3. Image of $$\exp$$ is a subset of $$G^0$$.

4. 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 $$\mathbb R^n$$.

5. 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)

• @EricWofsey Can you answer about the exponential map here also please? This question is intended actually for that question. I'm trying to prove differential of $\exp$ is $\gamma$ itself (assuming that is true, which is what that question is asking). – Ekhin Taylor R. Wilson Nov 12 '19 at 4:00
First of all, there is absolutely no reason to talk about $$T_eU$$. Since $$i_{*,e}$$ is surjective, $$\exp(i_{\{*,e\}}(T_eU))$$ is just the same as $$\exp(T_eG)$$. So your question is, must the image of the exponential map be contained in an arbitrary open neighborhood of $$e$$?
The answer is then obviously no, since the intersection of all open neighborhoods of $$e$$ is just $$\{e\}$$ itself, and the image of the exponential map is more than just $$\{e\}$$ (assuming $$n>0$$). The same holds if you restrict to $$U$$ that are homeomorphic to $$\mathbb{R}^n$$, since $$e$$ has arbitrarily small neighborhoods that are homeomorphic to $$\mathbb{R}^n$$.
For a very simple explicit example, if $$G=\mathbb{R}$$, then the exponential map is surjective, so any open interval around $$0$$ is an open neighborhood homeomorphic to $$\mathbb{R}$$ that does not contain the image of the exponential map.