Clarification on Diestel's proof of Tutte's Theorem.

Question

Tutte's Theorem states:

A graph $G$ has a 1-factor if and only if $\text{odd}(G −S) \leq |S|$ for all $S \subseteq V (G)$.

I believe I understand why the forward implication holds. Diestel then presents an argument for the "only if": find a "bad" set of vertices that violates the "only if". In order to simplify things, Diestel states:

We may assume that G is edge-maximal without a 1-factor.

From this, Diestel proceeds to state

Clearly, if G contains a bad set S then, by its edge-maximality and the trivial forward implication of the theorem,

(*) [a] all the components of $G−S$ are complete and [b] every vertex $s \in S$ is adjacent to all the vertices of $G−s$.

I believe I am missing something, because none of this seems at all obvious to me

1. I believe [a] holds because of the edge maximality of $G$. My gut feeling is that if it were not the case you could simply pad each component with extra edges and this would contradict the maximality of $G$. But I am not sure.
2. I do not see why [b] can be assumed any help clarify why [b] must hold would be appreciated.

Answer 1

[a] and [b] both follow from the same idea: if $G$ has a bad set $S$, what are all the edges we could add and still keep the set $S$ bad? Any such edges must be already there, because if $S$ remains bad, then $G$ remains without a $1$-factor, and $G$ must be edge-maximal without a $1$-factor.

We can:

1. Add all edges within a component of $G-S$. This will not change the number or order of components of $G-S$, so in particular it will not change $\text{odd}(G-S)$, so in particular it will not affect whether $\text{odd}(G-S) > |S|$.
2. Add all edges with at least one endpoint in $S$. These edges will not affect $G-S$ at all, so once again they will not affect whether $\text{odd}(G-S) > |S|$.

Once we've done the first step, [a] holds; once we've done the second step, [b] holds.