A homomorphism from a connected Lie group is determined by its differential at the identity

Fix the neighborhoods $V,U$ supplied by [F3]. If $g\in U$, then $g={\exp }_{G}X$ for a unique $X\in V$. By [F2] and the hypothesis on the differentials, $F_1(g)={\exp }_H(d(F_1{)}_{e}X)={\exp }_H(d(F_2{)}_{e}X)=F_2(g)$. Thus $F_1$ and $F_2$ agree on the open identity neighborhood $U$.

Let $K=\{g\in G:F_1(g)=F_2(g)\}$. The identity belongs to $K$. If $g,h\in K$, then [F4] gives $F_1(gh)=F_1(g)F_1(h)=F_2(g)F_2(h)=F_2(gh)$, and similarly $F_1(g^{-1})=F_1(g{)}^{-1}=F_2(g{)}^{-1}=F_2(g^{-1})$. Hence $K$ is a subgroup of $G$, and step 1.1 gives $U\subseteq K$.

For each $k\in K$, the translate $kU$ is open and is contained in $K$; conversely every $k\in K$ belongs to $kU$ because $e\in U$. Therefore $K={\bigcup }_{k\in K}kU$ is open. Every left coset $gK$ is then open as well. If $g\mathrm{\notin }K$, the coset $gK$ is disjoint from $K$: an element $x\in gK\cap K$ would imply $g=x{k}^{-1}\in K$. Hence $G\setminus K={\bigcup }_{g\mathrm{\notin }K}gK$ is open, so $K$ is also closed.

The set $K$ is nonempty because it contains $e$. If its complement were nonempty, step 3.1 would partition $G$ into the two nonempty clopen sets $K$ and $G\setminus K$, contradicting connectedness by [F5]. Thus $K=G$, which means $F_1=F_2$.

Lie groups are nonempty and boundaryless. In dimension zero a connected Lie group is a one-point discrete space, so the conclusion also follows directly; dimension one requires no change. There is no metric, degeneracy, or endpoint issue. The only choice use is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited through [F2] and [F3]. Fixing one supplied neighborhood pair and forming unions over already specified sets select no family of witnesses. No biconditional is asserted.