if I have a (topologically !) finite generated profinite group $G$ (so there exist a finite 1-convergent subset $X$ (a subset X is called 1-convergent if $X$ \ $X \cap N$ is finite for each $N \vartriangleleft _O G$) such that $G = \bar{<X>}$ (or equivalently $G = <X>N$ for each $N \vartriangleleft _O G$) and we have two morphisms of profinite groups $\phi, \psi: G \to H$ which coincide at the set $X$ how I can conclude that they coincide at the whole $G$?
The problem is that $G$ is topologically generated by $X$, so I can't argue like for group which is generated by $X$ (therefore $G = <X \cup X^{-1}$) in common sense.
To say that a subset $X$ topologically generates $G$ means that $\overline{\langle X\rangle}=G$. In general this is not the same as having $\langle\overline{X}\rangle=G$. If $\varphi,\psi:G\to H$ are continuous homomorphisms from $G$ to another profinite group $H$ (or any Hausdorff topological group $H$) which coincide on $X$, then they coincide on $\langle X\rangle$ because $\varphi,\psi$ are homomorphisms, and then by continuity they also coincide on $G=\overline{\langle X\rangle}$.