I'm studying a proof that $SL_n(\mathbb{Z}_p)$ is profinite, but I'm stuck in one point.
If $a$ is a matrix, then $a^{(i,j)}$ denotes the entries of $a$.
The idea is show that $$SL_n(\mathbb{Z}_p) \simeq \varprojlim_i SL_n(\mathbb{Z}/p^i\mathbb{Z}).$$ In order to show the isomorphism, the author uses the natural projection maps $\epsilon_m: \mathbb{Z}_p \to \mathbb{Z}/p^m\mathbb{Z}$ and defines the homomorphism $\alpha: SL_n(\mathbb{Z}_p) \to \varprojlim_i SL_n(\mathbb{Z}/p^i\mathbb{Z})$ by $g \mapsto (g_m)_{m \in \mathbb{Z}}$ where $g_m^{(i,j)} = \epsilon_m(g^{(i,j)})$. So, he claims that $\alpha$ is continuous because $$S_m = \prod_{i=1}^m \{1\} \times \prod_{i=m}^{\infty}SL_n(\mathbb{Z}/p^i\mathbb{Z})$$ form a base for the neighborhoods of the identity of inverse limit and $\alpha^{-1}(S_m) = SL_n(p^m\mathbb{Z}_p)$ is open in $SL_n(\mathbb{Z}_p)$.
I cannot see why $\alpha^{-1}(S_m) = SL_n(p^m\mathbb{Z}_p)$.
The proof can be found in this link.