Image of inertia in $\textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})^{\textrm{ab}}$

Question

Let $p$ be a prime. Choosing an embedding $\overline{\mathbb{Q}} \to \overline{\mathbb{Q}_p}$ induces an injection $\textrm{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p) \to \textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ and thus a homomorphism $\textrm{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)^{\textrm{ab}} \to \textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})^{\textrm{ab}}$. By class field theory, $$\textrm{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)^{\textrm{ab}} \cong \mathbb{Z}_p^{\times} \times \widehat{\mathbb{Z}},$$ where the $\mathbb{Z}_p^{\times}$ factor corresponds to the image of the inertia subgroup $I_p$, and $$\textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})^{\textrm{ab}} \cong \prod_{\ell} \mathbb{Z}_{\ell}^{\times},$$ where the product ranges over all primes $\ell$. My question is what is the composition $$\mathbb{Z}_p^{\times} \to \mathbb{Z}_p^{\times} \times \widehat{\mathbb{Z}} \cong \textrm{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)^{\textrm{ab}} \to \textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})^{\textrm{ab}} \cong \prod_{\ell} \mathbb{Z}_{\ell}^{\times} \to \mathbb{Z}_p^{\times},$$ where the first map is the natural inclusion and the last one is the projection onto the $\mathbb{Z}_p^{\times}$ factor. Is it just the identity on $\mathbb{Z}_p^{\times}$? This seems to make sense, as both sides consist of elements which fix prime-to-$p$ roots of unity and act on $p$-power roots of unity.