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Source. Theorem 1a, printed p. 150 (published PDF).

Statement. In the notation of the maximum-union formula, there are partial transversals $T_1,\ldots,T_k$ with $|T_t|=n_t$ and $\bigcup_tT_t=E$ if and only if

$$n_t\le\rho(E)\quad(1\le t\le k),\qquad |A|\le\sum_{j=1}^{\sigma(A)}n_j^* =\sum_t\min(n_t,\sigma(A)) \quad(A\subseteq E). \tag{1}$$

In this criterion $\rho(A)$ may replace $\sigma(A)$. The sets in the cover need not be disjoint.

Proof. A covering partial transversal has size at most $\rho(E)$. For each $A$, its intersection with $T_t$ has at most $n_t$ elements and uses at most $\sigma(A)$ distinct family indices. Therefore

$$|A|\le\sum_t|A\cap T_t| \le\sum_t\min(n_t,\sigma(A)).$$

This proves necessity. The same argument with the rank bound on $A\cap T_t$ also proves necessity of the rank-form criterion.

Conversely, (1) says that every objective in the maximum-union formula is at least $|E|$. That maximum cannot exceed $|E|$, so there is a disjoint packing of partial transversals $S_t$ of sizes at most $n_t$ whose union is $E$.

Partial transversals are independent in the transversal matroid. Extend each $S_t$, separately, to an independent set $T_t$ of size $n_t$, using $n_t\le\rho(E)$ and finite basis extension. The union remains $E$, though these extensions may overlap. This gives the required cover. The rank version follows by the same reasoning using the rank minimum formula. $\square$