In the proof of $5$-colorable planar graph $G$.
We have the base case $n\le5$ then the induction case for $n+1\ge6$. We want to show that $G$ is $5$-colorable. I will not detail the proof for base case.
Induction hypothesis: Let us assume that every simple, planar graph on $n$ vertices where $n\ge5$ is $5$-colorable.
In one case out of many cases that follows from the induction hypothesis, we might have vertex $u$ where all of its 5 neighbors take different colors, say $(1,2,3,4,5)$. For this particular case that I am asking about, we might consider the subgraph $H$ of $G$ consisting vertices colored either 1 and 3 and all edges between these vertices (subgraph spanned by these vertices) given that we have vertices $(a,b,c,d,e)$ colored $(1,2,3,4,5)$ respectively. Vertices $a$ and $c$ are in $H$ because they get colored $1$ and $3$.
My problem: Since $H$ is $2$-colorable, then it's bipartite. In the proof, it's mentioned that if $H$ has several components and $a$ and $c$ lie in different components, then swap the color of vertices in the component containing $a$. This gives another $5$-coloring of $G-u$ in which the color of $a$ switch to 3 but the color of $c$ remain 3, but now we can color $u$ with 1 to get a $5$-coloring of $G$ (if we get lucky). Can you please clarify my problem in another words as I don't understand how $a$ and $c$ lie in different components while we choose them to be in a bipartite graph with 2 colors?