Let $G_1,...,G_n$ be groups and $T:=G_1× \cdots × G_n$.
Show that $ \forall i \in \{1, \cdots , n \} $ exists an epimorphism $\phi_i: T \rightarrow G_i$ and find $\ker(\phi_i)$.
I'm not sure if what I have done is correct:
Let $\phi_i (g_1, \cdots , g_n) = g_i $ for $i\in \{ 1, \cdots , n\}$
We want to see first if $\phi$ is a homomorphism, so let's consider:
$(g_1, \cdots , g_n),(h_1, \cdots , h_n) \in T$
then
$\phi_i ((g_1, \cdots , g_n) \cdot (h_1, \cdots , h_n)) = \phi_i ((g_1h_1, \cdots , g_nh_n)) = g_ih_i$
$\Rightarrow \phi_i (g_1, \cdots , g_n) \phi_i(h_1, \cdots , h_n) = g_ih_i $
And now we want to see $\phi_i$ is surjective. Consider $g_i \in G_i$
$\Rightarrow \phi_i (e, \cdots , g_i , \cdots, e) = g_i$
And since then, $\phi_i$ is surjective. On the other hand
$\ker(\phi_i)=\{ (g_1, \cdots , g_{i-1}, e , g_{i+1}, \cdots , g_n) | g_i \in G_i \}$
Yes, your solution is fine.
These homomorphisms are natural consequences of the universal property of the direct product.