Are all anabelian Galois actions faithful?

Question

Let $C/\mathbb Q$ be a smooth projective curve of genus $g\geq 2$ or a smooth affine curve of genus $g \geq 1$. The exact sequence

$1 \to \pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q}) \to \pi_1^{et}(C) \to \operatorname{Gal}(\bar{\mathbb Q}|\mathbb Q) \to 1$

gives a homomorphism from $\operatorname{Gal}(\bar{\mathbb Q}|\mathbb Q)$ to the outer automorphism group of $\pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q})$.

Is this homomorphism always injective?

If one instead takes $C$ to be a curve of genus $0$ with $3$ points removed, it is injective, by Belyi's Theorem.

Answer 1

The answer is "yes" I think, even if you replace $\mathbb Q$ with a number field.

In the affine case this is a result of Matsumoto, as pointed out by Felipe Voloch, see

Matsumoto, Makoto Galois representations on profinite braid groups on curves. J. Reine Angew. Math. 474 (1996), 169–219.

In the proper case this is a more recent result of Hoshi and Mochizuki, see

Hoshi, Yuichiro; Mochizuki, Shinichi On the combinatorial anabelian geometry of nodally nondegenerate outer representations. Hiroshima Math. J. 41 (2011), no. 3, 275–342.

See also Tamás Szamuely's nice survey:

Heidelberg lectures on fundamental groups, in J. Stix (ed.) The Arithmetic of Fundamental Groups (PIA 2010), Contributions in Mathematical and Computational Sciences, Vol. 2, Springer-Verlag, 2012, 53--73

available here http://www.renyi.hu/~szamuely/pia.pdf

Answer 2

In the affine case, when the divisor at infinity has at least one rational point, this follows from theorems 2.1 and 2.2 of A. Matsumoto, A. Tamagawa, Mapping-class-group action versus Galois action on profinite fundamental groups. Amer. J. Math. 122 (2000), 1017--1026. I don't know what happens in the general case.