I am trying to understand the one line proof of Tutte-Berge formula here in 24.1 (just the $\le$ side proof), i.e. $$ν(G) ≤ |U| + ν(G − U) ≤ |U| + \frac 1 2 (|V \setminus U| − o(G − U)) = \frac 1 2 (|V | + |U| − o(G − U))$$
I do not fully understand where the first two inequalities and the third equality come from. Any explanation would be appreciated, thanks.
For the first inequality $$\nu(G)\leq |U|+ \nu(G-U),$$ take any matching in $G$ and split it into edges that contain an element of $U$ and edges that do not. There are at most $|U|$ edges of the first type, and at most $\nu(G-U)$ edges of the second type since they form a matching in $G-U$.
The second inequality $$\nu(G-U)\leq \frac{1}{2}(|V\setminus U|-o(G-U))$$ is because any matching in a graph must omit a vertex from each odd component. So, a matching in $G-U$ can involve at most $|V\setminus U|-o(G-U)$ vertices.
The last equality is just algebra since $|V\setminus U|=|V|-|U|$.