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Let $S$ be constructible in an algebraic group, and $C=\overline{S}$. The closure is a subgroup. Choose a dense open $U\subseteq C$ contained in $S$. For every $c\in C$, the dense opens $U,cU$ intersect, giving $u=cv$ for $u,v\in S$. Hence $c=u{v}^{-1}\in S$ and $S=C$. The Chevalley constructibility theorem therefore proves closedness of homomorphism images.

The dimension of a topological space by irreducible chains is the supremum of the integers $d$ for which there is a chain $Z_0⊊Z_1⊊\cdots ⊊Z_d$ of nonempty irreducible closed subsets of the space. Closedness here is relative to the given locally closed space. For varieties this is their Krull dimension.

An algebraic group is a group whose underlying space is an algebraic variety and whose multiplication $G\times G\to G$ and inversion $G\to G$ are morphisms. An algebraic group action on a variety $V$ is a morphism $G\times V\to V$ satisfying the identity and associativity axioms of a group action.

For the homomorphism $\phi$, the kernel is ${\phi }^{-1}(1_H)$, so the kernel is closed. To prove closedness of the image, use the Chevalley constructibility theorem: the image of a morphism of varieties is constructible. Thus $S=\phi (G)$ is a constructible subset of a variety and an abstract subgroup. Its closure $C$ is also a subgroup: translation by elements of $S$ preserves $C$, and continuity then extends multiplication and inversion to $C$.

A dense constructible subset contains a dense open subset $U$ of its closure. For $c\in C$, both $U$ and $cU$ are dense open subsets of $C$, so their intersection is nonempty. If $u=cv$ with $u,v\in U\subseteq S$, then $c=u{v}^{-1}\in S$. Hence $S=C$, proving that the image is closed. This is the principle that a constructible subgroup is closed.

Every nonempty fiber of $G\to S$ is a translate of $\ker \phi$ and has that same dimension. The fiber dimension theorem therefore givesThis dimension formula for an algebraic group homomorphism is a dimension statement, so it does not require separability of $\phi$.

For a dimension vector of a quiver representation $\mathbf{n}$, setThe base change action on quiver representations isThe entries are regular functions on the product of the general linear groups and the quiver representation space, because inverse entries are cofactors divided by the invertible determinant. Thus this is an algebraic group action.

For nonzero $\mathbf{n}$, let $\Delta k^{\times }$ be the common scalar subgroup $(\lambda I_{n_i}{)}_i$ and define $\mathrm{PGL}(\mathbf{n})=\mathrm{GL}(\mathbf{n})/\Delta k^{\times }$. Common scalars act trivially, so the formula descends to an algebraic group action of this projective base change group of a quiver. This is a quotient by one common scalar, not a product of the individual projective groups. If every $n_i=0$, the representation space is a point and both actions are taken to be trivial.