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Source. Ford–Fulkerson (1958), Corollary and equation (13), printed p. 83 (published scan).

Two indexed families $\mathcal S,\mathcal T$ of $n$ subsets of a finite ground set have a common SDR if and only if

$$|X|+|Y|\le n+|I_{\mathcal S}(X)\cap I_{\mathcal T}(Y)| \qquad(X,Y\subseteq[n]).$$

A common SDR means two injective representative assignments with the same range, permitting different assignments of that range to the two families' indices.

Proof. In Theorem 2, take $\alpha_i=0$ and $\beta_i=1$ for every element. An SRR is then injective. Equal multiplicities mean exactly equal ranges, and the theorem's weighted intersection sum becomes the cardinality in the displayed criterion. Its equivalence proves both directions, including the empty-family case. $\square$

Bears on. Common finite transversals. This is more than the separate Hall conditions for each family: the intersection term couples their possible choices of a shared range. No Erdős problem: the paper states no relation to a numbered Erdős problem.

Prescribed-multiplicity comparison. Welsh (1969), Theorem 12 allows nonnegative multiplicities $p_i,q_i$ with the same total $N$. Setting $p_i=q_i=1$ makes $N=n$ and turns its criterion into

$$|I_{\mathcal S}(X)\cap I_{\mathcal T}(Y)|\ge |X|+|Y|-n,$$

after identifying ground elements with their indices. Rearranging gives the Ford–Fulkerson common-SDR inequality displayed above. This is an exact specialization of the Welsh statement, not a second proof of this 1958 corollary.