Perfect matching in a bipartite graph avoid increasing perfect matchings?

Question

Let $A=\{a_1,\dots,a_n\}$ and $B=\{b_1,\dots, b_n\}$ be two disjoint vertex sets. Let $A_i=\{a_1,\dots, a_i\}$ be the set of first $i$ vertices of $A$. And $B_i=\{b_1,\dots, b_i\}$, similarly.

For $1\le i_1\le i_2\le\dots\le i_k\le n$, suppose we have $M_j$ that is a perfect matching between $A_{i_j}$ and $B_{i_j}$ for each $1\le j\le k$ satisfying $$M_j\cap(\cup_{1\le \ell<j}M_{\ell})=\emptyset,$$ i.e., those $M_j$ are edge-disjoint.

Suppose the degree of each vertex of $A_{i_k}$ in $\cup_{1\le j\le k}M_j$ is strictly less than $i_k$. Then is it true that there is still a perfect matching between $A$ and $B$ that is edge-disjoint with $\cup_{1\le j\le k}M_j$?

I tried some small examples where the statement is true.

Answer 1

It is false: let $M_1=\{a_1b_1,a_2b_2\}$ and $M_2=\{a_1b_2,a_2b_1,a_3b_3\}$ (so that $i_1=2,i_2=3$). If $n=3$, there is no perfect matching out of $M_1\cup M_2$, although each vertex has degree at most 2, which is less than 3.

Answer 2

No, that is false. Consider $i_1 = i_2 = 2$ and $i_3 = 3$. Then $\{\,M_1, M_2\,\} = \{\,\{\,a_1b_1, a_2b_2\}, \{\,a_1b_2, a_2b_1\,\}\,\}$. And $M_3$ should contain at least one of these $4$ edges to be a perfect matching.