*I'm trying to make the proof of Yamabe's Theorem that says that an arcwise connected subgroup of a Lie group G is a Lie subgroup of G. I found the proof in Goto's article (https://www.ams.org/journals/proc/1969-020-01/S0002-9939-1969-0233923-X/S0002-9939-1969-0233923-X.pdf) and in the book "Structure and Geometry of Lie Groups", from Joachim Hilgert and Karl-Hermann Neeb. In booth of them, they use the following lemma to conclude the proof of Yamabe's theorem:
Blockquote: Lemma: If $H$ is an arcwise connected topological subgroup of $G$, then for each identity neighborhood $U$ of $1$ in $G$, the arc-component $U_a$ of $U\cap H$ generates $H$.
On the first one, Goto just says that the proof of this lemma is trivial. On the second one, they make a proof:
Blockquote: Let $H_U ⊆ H$ denote the subgroup generated by $U_a$. Further, let $h\in H$ and $\gamma : [0,1] \to H$ be a continuous path from $1$ to $h$. We consider the set $$S := \{t ∈ [0, 1] : γ(t) ∈ H_U\}.$$ Then the subset $\gamma^{−1}(U)$ of $S$ is a neighborhood of zero. For any $t_0 ∈ S$, there exists an $\epsilon > 0$ with $γ(t_0)^{−1}γ(t) ∈ U_a$ for $|t−t_0| ≤ ε$. This implies that $S$ is an open subset of $[0, 1]$. If $t_1 \in [0, 1]\setminus S$, then we also find an $\epsilon > 0$ with $\gamma(t_1)^{−1}\gamma(t) \in U_a \subseteq H_U$ for $|t − t_1| ≤ \epsilon$, so that $\gamma(t_1) ∈ H_U$ leads to $\gamma(t) ∈ H_U$ for $|t − t_1| ≤ \epsilon$. Thus $[0, 1] \setminus S$ is also open in $[0, 1]$, and now the connectedness of $[0, 1]$ implies that $S = [0, 1]$, hence $h = \gamma(1) \in H_U$.
Apparently, where I emphasis added, they used that $U_a$ is a neighborhood of $1$ in $H$. But how can I guarantee this if I don't have that $H$ is a locally connected space? Is there some result that makes this real for topological subgroups of a Lie group? I tryed to proof that but I coudn't. I'd be grateful if anyone had any tips.*
You are right: $U_a$ is not necessarily a neighbourhood of $1$ in $H$, and $H$ is not necessarily locally connected.
However, since $U$ is open, for any $t_0\in S$ there exists $\varepsilon >0$ such that for all $t\in [0,1]$ with $|t-t_0|<\varepsilon$ one has $\gamma(t_0)^{-1}\gamma(t)\in U$. Since the image of $\gamma$ lies in $H$, we also have $\gamma(t_0)^{-1}\gamma(t)\in U\cap H$. It is, moreover, in the same arc-component (in $U\cap H$) as the element $1$ (since it is connected to it by the path $$s\mapsto \gamma(s)^{-1}\gamma(t)$$ where $s$ goes from $t$ to $t_0$), so $\gamma(t_0)^{-1}\gamma(t)$ is in $U_a$.