Let $\mathrm{G=N_1 \times N_2}$ be an internal direct product of groups and let $\phi:G \to H$ be a surjective homorphism. Is it true that we also have the internal direct product $\mathrm{H= \phi(N_1) \times \phi(N_2)}$?
I have shown that $H=\phi(N_1)\phi(N_2)$ and that $\mathrm{ϕ(N1); \ ϕ(N2)}$ are normal subgroups of $H$ and if the statement above is true, I only need to prove that $\phi(N_1) \cap \phi(N_2)$ = $\{e\}$. Any suggestions to prove or disprove this last step?
This is false. Consider $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ by $\phi(x,y)=x-y$. This is a surjective homomorphism. However, $\phi(\mathbb{Z} \times \{0\})= \mathbb{Z}$ and $\phi(\{0\} \times \mathbb{Z})=\mathbb{Z}$ and $\mathbb{Z}$ cannot be the direct product of two copies of itself.
Note that this counterexample works because the kernel of $\phi$ is large. Your desired statement is true when $\phi$ is an isomorphism, so any counterexample needs a non-trivial kernel.
The "internal-ness" doesn't really matter since $\mathbb{Z} \times \mathbb{Z}$ is the internal direct product of $\mathbb{Z} \times \{0\}$ and $\{0\} \times \mathbb{Z}$.