Diestel writes in his Graph Theory book (Proposition 1.7.1) that
The minor relation and the topological-minor relation are partial orderings on the class of finite graphs, i.e. they are reflexive, antisymmetric and transitive
I proved the three properties for the minor relation but am confused when it comes to the topological-minor relation and proving that it is reflexive. Diestel does not really define if a graph $G$ is counted as a subdividing itself, i.e. having an empty set of subdividing vertices and reading the definition on Proof Wiki sounds like one has to at least subdivide one edge resulting in a non-isomorphic graph $G'$ such that $G'$ is the smallest subdivision of $G$.
However, to prove reflexivity, i.e. $G$ being a topological minor of $G$, $G$ must be in the set of subdividing graphs $TX$. This is due to the fact that the topolocical minor relation is defined as follows: $X$ is a topological minor of $Y$ if there exists a subdivision of X, which is a subgraph of $Y$. Therefore, if $G\notin TX$ because $G'$ is the smallest subdivision of $G$. Then, no graph in $TX$ can be a subgraph of $G$ and hence, $G$ can't be its own topological minor.
Am I misunderstanding how to prove the reflexivity of a topological minor or something in its definition?
PS: This post asked a similar but different question.
We want the relations to be reflexive, so we define a graph $G$ to be a subdivision of itself.
You could play around with definitions you found in Diestel or on Proof Wiki until this works. For example, Proof Wiki says that $G'$ is a subdivision of $G$ if it is obtained from $G$ by a sequence of edge subdivision operations, so maybe we allow the empty sequence.
But that's missing the point. Definitions are not set in stone; we choose our definitions so that it's easy to say what we want to say with them. It causes multiple problems if we don't allow a graph to be a subdivision of itself. For example, Kuratowski's theorem would have to be changed to "A graph is planar if and only if it does not contain a subdivision of $K_5$ or $K_{3,3}$ and also is not isomorphic to $K_5$ or $K_{3,3}$" which is awkward.
You'll notice that we can still state the theorem even if we make the wrong definitional choice. Similarly, if we said that $G$ is not a subdivision of itself, we could say that the subdivision relation is a strict partial order, instead. The math doesn't change; only how we talk about it does. It's important to choose definitions to make it easiest to talk about the math.