Let $G_1,...,G_n$ be groups. Show that there exist an injective morphism $\xi:$$\prod_{i=1}^{n}\text{Aut}(G_i)\to \text{Aut}\Big(\prod_{i=1}^{n}G_i\Big)$. I would like to know if my proof holds, please. And if not, to have an explication.
Let $\phi_i \in \text{Aut}(G_i)$ for $i=1,...,n$ and $(g_1,...,g_n) \in \prod_{i=1}^{n}G_i$. Define $\xi$ as the following:
$\xi(\phi_1,...,\phi_n)(g_1,...,g_n)=(\phi_1(g_1),...,\phi_n(g_n))$.
One can check that $\xi$ is a group morphism.
We check now that $\xi$ is injective:
Let $\xi(\phi_1,...,\phi_n)(g_1,...,g_n)=\xi(\phi'_1,...,\phi'_n)(g_1,...,g_n)$. We show now that $\phi_i=\phi'_i$ for $i=1,...,n$:
$\xi(\phi_1,...,\phi_n)(g_1,...,g_n)=\xi(\phi'_1,...,\phi'_n)(g_1,...,g_n)\implies (\phi_1(g_1),...,\phi_n(g_n))=(\phi'_1(g_1),...,\phi'_n(g_n)) \implies \phi_i=\phi'_i$ for $i=1,...n$ so $\xi$ is injective.