homomorphism between smooth algebraic groups of the same dimension For Lie groups, we have a theorem:

Suppose $G$ and $G'$ are Lie groups of the same dimension, $G'$ is connected, and $f : G \to G'$ is a homomorphism of Lie groups with discrete kernel. Then, $f$ is surjective. (In particular, $G$ is a covering of $G'$.)

Clearly, this also works for algebraic groups in characteristic $0$. My question is does a similar statement hold for smooth algebraic groups in arbitrary characteristic?
 A: Yes, this is true. That said, the covering part is not in general true since $f$ needn't be faithfully flat (I'll give an example of this later). So, let's assume that $G'$ is reduced.
Indeed, note that $\ker(f)$ being a closed subscheme of $G$ which is Noetherian is finite. Moreover, we have that we have a factorization
$$G\to G/\ker(f)\hookrightarrow G'$$
where the first map is the canonical quotient map and the second map is a closed immersion (e.g. see [Mil, Theorem 5.39]). Note though that since $\ker(f)$ is finite we have that $G/\ker (f)$ has the same dimension as $G$ (e.g. see [Mil, Proposition 5.23]) and thus $G/\ker(f)\hookrightarrow G'$ is a closed immersion where both objects have the same dimension. This implies, first of all, that the image of $G/\ker(f)$ has topological image an irreducible component of $G'$ which, since $G'$ is connected and thus irreducible (e.g. see [Mil, Corollary 1.35]), this implies that $G/\ker (f)$ has the same underlying set of $G'$. Since $G'$ is reduced this implies that $G/\ker (f)\to G'$ is an isomorphism and thus $G'\cong G/\ker (f)$ so that $G$ covers $G'$.
Note though that the inclusion $\{e\}\hookrightarrow \mu_p$ (where $\{e\}$ denotes the trivial group scheme) is an injective map where both have the same dimension, but is not a covering (in the sense that it's a quotient map with finite central kernel).
[Mil] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.
