What is the kernel of the map $Out(\widehat{F_2})\rightarrow GL_2(\widehat{\mathbb{Z}})$?

Let $F_2$ be the free group on two generators, and $\widehat{F_2}$ its profinite completion. Let $Out(\widehat{F_2})$ be the outer automorphism group of $\widehat{F_2}$, ie, $Out(\widehat{F_2}) = Aut(\widehat{F_2})/Inn(\widehat{F_2})$. There's a natural surjection $$Aut(\widehat{F_2})\rightarrow GL_2(\widehat{\mathbb{Z}})$$ Certainly $Inn(\widehat{F_2})$ is in the kernel, so we get the map $Out(\widehat{F_2})\rightarrow GL_2(\widehat{\mathbb{Z}})$.

Furthermore, since $Out(F_2)\cong GL_2(\widehat{\mathbb{Z}})$ is known to have the congruence subgroup property, we have an injection $\widehat{Out(F_2)}\rightarrow Out(\widehat{F_2})$.

Let $G\Gamma(N)$ for $N\ge 1$ denote the congruence subgroup of $GL_2(\mathbb{Z})$ of level $N$, then we also have an exact sequence $$1\longrightarrow\bigcap\overline{G\Gamma(N)}\longrightarrow\widehat{GL_2(\mathbb{Z})}\longrightarrow GL_2(\widehat{\mathbb{Z}})$$ where $\overline{G\Gamma(N)}$ denotes the closure inside $\widehat{GL_2(\mathbb{Z})}$. Here, $\bigcap\overline{G\Gamma(N)}$ is the congruence kernel of $GL_2(\mathbb{Z})$ and is rather big (contains as a subgroup of index 2, the congruence kernel of $SL_2(\mathbb{Z})$ which is a free profinite group of countably infinite rank)

Thus, $\bigcap\overline{G\Gamma(N)}$ is also in the kernel of $Out(\widehat{F_2})\rightarrow GL_2(\widehat{\mathbb{Z}})$.

• Since $Out(F_2)$ is known to have the congruence subgroup property, whereas $GL_2(\mathbb{Z})$ doesn't, the kernel will contain the elements of $\widehat{Out(F_2)}\cong\widehat{GL_2(\mathbb{Z})}$ in the kernel of $\widehat{GL_2(\mathbb{Z})}\to GL_2(\widehat{\mathbb{Z}})$. – HJRW Apr 29 '15 at 20:23
• Rephrasing the first comment: $Aut(F_2)$ is known to have the subgroup congruence property, whereas $Out(F_2) \simeq GL_2(\mathbb{Z})$ doesn't. See "The congruence subgroup property for $Aut(F_2)$: a group-theoretic proof of Asada's theorem" by K. Bux, M. V. Hershov and A. S. Rapinchuck, 2011. Regarding this part, the question needs to be edited. – Luc Guyot Jun 19 '16 at 19:41