Let $G$ be a linear algebraic group over an algebraically closed field $k$. Consider closed subgroups $N \subseteq H \subseteq G$ such that $N$ is a normal subgroup of $G$. Then restricting the quotient $$\pi_G: G \to G/N$$ to $H$ and applying the universal property of quotients gives us a morphism $$\varphi : H/N \to G/N$$ of algebraic groups such that $\varphi \circ \pi_H = \pi_G \circ \iota$ where $\pi_H : H \to H/N$ and $\iota: H \to G$ is the inclusion.
Question: Is $\varphi$ an isomorphism onto its image?
I believe this to be true, but when translating the situation to rings of regular functions I would have to show that restriction $k[G]^N \to k[H]^N$ of $N$-invariant regular functions is surjective and I do not see why this is true. I read in Humphrey's book on linear algebraic groups that it suffices to show that $\varphi$ or $\pi_G \circ \iota : H \to \pi_G(H)$ is separable but I do not know how to do this either (my knowledge on algebraic geometry is quite limited).
Any help is appreciated.
I believe I have found a result that answers my question positively.
In Borel's 'Linear Algebraic Groups', Proposition 6.7 in Chapter II says, that given a $k$-group $G$ acting $k$-morphically on a $k$-variety $V$, then for $x \in V(k)$ and $G(x)$ the orbit of $x$ under $G$, the map $\psi: G \to G(x), g \mapsto g \cdot x$ is a quotient map for the stabilizer $G_x$ of $x$ if and only if $\ker(\text{d}\psi_e)$ is contained in $L(G_x)$, the Lie algebra of $G_x$.
In our situation $H$ acts on $G/N$ and taking $x = e \in G/N$ we have $\psi = \pi_G \circ \iota$ and $G_x = N$. We have $\ker((\text{d}\pi_G)_e) = L(N)$ and so $$\ker(\text{d}\psi_e) = \ker((\text{d}\pi_G)_e \circ (\text{d}\iota)_e) = L(H) \cap L(N) = L(N) = L(G_x)$$ because $N \subseteq H$. Consequently, $\pi_G \circ \iota : H \to \pi_G(H)$ is a quotient morphism for $N$ as desired.