In Diestel's Graph Theory book he gives a proof of Kruskal's Theorem. I am trying to go through the details to make sure I fully understand.
Kruskal's Theorem The finite trees are well-quasi-ordered by the topological minor relation.
Before the proof even begins, we define a new relation, (instead of using the topological minor relation). I've copied the paragraph below which comes before the proof starts.
"We shall base the proof ... on the following notion of an embedding between rooted trees, which strengthens the usual embedding as a topological minor. Consider two trees $T$ and $T'$, with roots $r$ and $r'$ say. Let us write $T \leq T'$ if there exists an isomorphism $\varphi$, from some subdivision of $T$ to a subtree $T''$ of $T'$, that preserves the tree-order on $V(T)$ associated with $T$ and $r$. (Thus is $x < y$ in $T$ then $\varphi(x) < \varphi(y)$ in $T'$) As one easily checks, this is a quasi-ordering on the class of all rooted trees."
The topological minor relation says that $T \leq T'$ if a graph isomorphic to $T$ is a topological minor of $T'$, i.e. $T'$ contains a subdivision of a graph isomorphic to $T$.
Question 1. How is this "new" relation (described in quotations) stronger? When I read it, it sounds like a different way to describe the topological minor relation. Is it because of the preservation of the tree-order? Or because we are restricting it to only rooted trees?
Question 2. Diestel states that you can easily check this is a quasi-ordering on the rooted trees. I want to make sure I understand this as well. A quasi-ordering must be reflexive and transitive, (not necessarily antisymmetric).
For reflexive, can we let $\varphi$ be the identity? This is an isomorphism from $T$ to itself. Clearly this preserves tree-order and we will get $T \leq T$, for any $T$ in the class of rooted trees.
For transitive, we know $\exists \varphi_1$ an isomorphism from a subdivision of $T_1$ to a subtree $T_2'$ of $T_2$, and $\varphi_2$ an isomorphism from a subdivision of $T_2$ to a subtree $T_3'$ of $T_3$, (tree-order preserving). So that $T_1 \leq T_2$ and $T_2 \leq T_3$. Now, we need to find a $\varphi$ satisfying the conditions so that $T_1 \leq T_3$. Can we let $\varphi = \varphi_2 \circ \varphi_1$? Then $\varphi$ is an isomorphism from the same subdivision of $T_1$ to the same subtree $T_3'$ of $T_3$.
My "question" is if my thinking is correct.
I would greatly appreciate any help/explanation/comments!