Show that $\prod_{i=1}^{n}\text{Aut}(G_i)\to \text{Aut}\Big(\prod_{i=1}^{n}G_i\Big)$ is injective

Question

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.

Answer 1

Your approach is broadly correct (and your definition of $\xi$ is the correct definition), but here is one minor-but-important point you are missing: you need to say "for all $g_i\in G_i$" somewhere. That is, if $\phi_i(g_i)=\phi_i'(g_i)$ then we cannot conclude that $\phi_i=\phi_i'$. Instead, we need this equality to hold for all $g_i\in G_i$.

So something like:

Suppose $\xi(\phi_1,\ldots,\phi_n)(g_1,\ldots,g_n)=\xi(\phi'_1,\ldots,\phi'_n)(g_1,\ldots,g_n)$ holds for all $(g_1, \ldots, g_n)\in G_1\times\cdots\times G_n$. We show now that $\phi_i=\phi'_i$ for $i=1,\ldots,n$:

\begin{align*}\xi(\phi_1,\ldots,\phi_n)(g_1,\ldots,g_n)&=\xi(\phi'_1,\ldots,\phi'_n)(g_1,\ldots,g_n)\\\implies (\phi_1(g_1),\ldots,\phi_n(g_n))&=(\phi'_1(g_1),\ldots,\phi'_n(g_n))\\ \Rightarrow\phi_i(g_i)&=\phi_i'(g_i)&\forall i.\end{align*} As the equality "$\phi_i(g_i)=\phi_i'(g_i)$" holds for all $g_i\in G_i$, it follows that $\phi_i=\phi'_i$ for $i=1,...n$ so $\ker\xi$ is trivial, so $\xi$ is injective.