What is the $4k$ kernelization algorithm for Planar Independent Set?
The formal problem is:


Input: A planar graph $G = (V, E)$ and an integer $k$ (the parameter)
Question: Does $G$ have an independent set of size at least $k$?


A kernelization algorithm for Planar IS with $4k$ vertices.

Recall that we can obtain a four-coloring of $G$ in polynomial time.  That is, we get a partitioning of your vertex set $V$ into $V_1, V_2, V_3, V_4$ such that each class is an independent set.

We now distinguish between two cases:


One class $V_i$ has size at least $k$.  This is a witness and hence $G,k$ is a yes instance, so we may output a trivial yes instance.
Otherwise, every class has size less than $k$, so $V$ has at most $4k-4$ vertices.


Thus in polynomial time, we may output a graph $G'$ with at most $4k$ vertices such that $G,k$ is a yes instance if and only if $G', k$ is a yes instance.

Notice that this only holds for planar graphs.  Independent Set is W[1]-hard on general graphs.