[proofplan] We verify the independent-set axioms for $\mathcal I$. The empty set and hereditary closure are immediate from deleting paths in a linkage. The exchange axiom is the main point: given linkages from $I$ and $J$ to $T$ with $|I| < |J|$, we fix a linkage of $I$, build the usual vertex-capacitated residual digraph, and compare it with a linkage of $J$. The signed difference of the two split linkages decomposes into alternating directed walks; since $J$ has more sources than $I$, at least one such walk starts at a source in $J \setminus I$ and ends at an unused terminal arc. That walk is an augmenting residual route, and toggling along it gives a linkage of $I \cup \{x\}$ for some $x \in J \setminus I$. [/proofplan]

[step:Verify the empty set and hereditary axiom] The empty set is linkable to $T$ by the empty family of directed paths, so $\varnothing \in \mathcal I$. Now let $I \in \mathcal I$, and let $I_0 \subseteq I$. Choose a linkage $(P_i)_{i \in I}$ from $I$ to $T$. The subfamily $(P_i)_{i \in I_0}$ is still pairwise vertex-disjoint, each path starts at its indexing vertex, and its endpoints in $T$ remain distinct. Hence $I_0$ is linkable to $T$, so $I_0 \in \mathcal I$. [/step]

[step:Construct the split residual digraph for a fixed linkage of $I$] Let $I,J \in \mathcal I$ satisfy $|I| < |J|$. Fix a linkage \begin{align*} \mathcal P = (P_i)_{i \in I} \end{align*} from $I$ to $T$. For each $i \in I$, write $t_i \in T$ for the terminal vertex of $P_i$, and define \begin{align*} T_{\mathcal P} = \{t_i : i \in I\}. \end{align*} We form a split directed graph $N_{\mathcal P}$. For each vertex $v \in V$, introduce two vertices $v^-$ and $v^+$ and a capacity arc \begin{align*} c_v := (v^-,v^+). \end{align*} For each directed arc $(u,v) \in A$, introduce the transit arc \begin{align*} a_{uv} := (u^+,v^-). \end{align*} Finally, introduce a new sink vertex $\omega$ and, for each terminal $t \in T$, a terminal arc \begin{align*} q_t := (t^+,\omega). \end{align*} The linkage $\mathcal P$ determines a family of arc-disjoint directed paths in $N_{\mathcal P}$ from the vertices $i^-$, with $i \in I$, to $\omega$: along a directed path \begin{align*} P_i = (v_0,v_1,\dots,v_m) \end{align*} with $v_0=i$ and $v_m=t_i$, the corresponding split path uses the arcs \begin{align*} c_{v_0}, a_{v_0v_1}, c_{v_1}, a_{v_1v_2}, \dots, a_{v_{m-1}v_m}, c_{v_m}, q_{v_m}. \end{align*} For a length-zero path at a terminal $i \in T$, the split path uses $c_i$ and $q_i$. Let $F_{\mathcal P}$ denote the set of all split arcs used by these paths. Define the residual digraph $R_{\mathcal P}$ on the same vertex set as $N_{\mathcal P}$ as follows: every arc of $N_{\mathcal P} \setminus F_{\mathcal P}$ is kept with its original orientation, while every arc of $F_{\mathcal P}$ is replaced by the reverse arc. Thus unused capacity, transit, and terminal arcs may be used forward, and used arcs may be cancelled by traversing them backward. Define the residual starting map \begin{align*} s_{\mathcal P}: V \to \{v^- : v \in V\} \end{align*} by $s_{\mathcal P}(x)=x^-$ for every $x \in V$. We always start at the incoming split vertex $x^-$; if $c_x$ is already used by the present linkage, the residual digraph records this by making the reverse arc $(x^+,x^-)$ available, so any rerouting through $x$ is handled by the ordinary residual-network operation rather than by changing the source vertex. [/step]

[step:Show that a residual route from $J \setminus I$ augments the linkage of $I$]Suppose there exists $x \in J \setminus I$ and a directed path $Q$ in $R_{\mathcal P}$ from $x^-$ to $\omega$. We claim that $I \cup \{x\}$ is linkable to $T$. Introduce a temporary source vertex $\rho$ and temporary source arcs $e_y=(\rho,y^-)$ for $y \in I \cup \{x\}$. Let $f_{\mathcal P}$ be the unit integral flow in the resulting network that uses the arcs $e_i$ for $i \in I$ and the split paths determined by $\mathcal P$. Extend $Q$ to a residual path from $\rho$ to $\omega$ by placing the unused arc $e_x$ before it. Augmenting the integral flow $f_{\mathcal P}$ by one unit along this residual path means adding every forward residual arc on the path and deleting the original arc corresponding to every backward residual arc on the path. At each internal vertex of the residual path, one unit of inflow and one unit of outflow are changed, so flow conservation is preserved. At $\rho$ the value increases by one, and at $\omega$ the value increases by one. The new flow is still integral and still takes only the values $0$ and $1$ on every arc, because forward residual arcs are precisely unused arcs and backward residual arcs are precisely used arcs. Since the only source arcs carrying flow after augmentation are $e_y$ with $y \in I \cup \{x\}$, the positive-flow arcs decompose, after deleting directed cycles if any occur, into pairwise arc-disjoint directed paths from $y^-$ to $\omega$ for all $y \in I \cup \{x\}$. The value is $|I|+1$, and there are exactly $|I|+1$ source arcs available, so each $y \in I \cup \{x\}$ supplies one path. The terminal arcs entering $\omega$ are distinct, because every terminal arc has capacity one in the split graph. Project these split paths back to $D$ by replacing each segment $u^+ \to v^-$ with the original directed arc $(u,v) \in A$ and by replacing each used capacity arc $v^- \to v^+$ with the vertex $v$. Arc-disjointness of the capacity arcs $c_v$ implies vertex-disjointness of the projected directed paths in $D$. The distinct terminal arcs $q_t$ imply that the projected endpoints in $T$ are distinct. Therefore the projected family is a linkage from $I \cup \{x\}$ to $T$, so $I \cup \{x\} \in \mathcal I$.[/step]

[guided]Assume that some $x \in J \setminus I$ has a residual directed path $Q$ from $x^-$ to $\omega$. To make the augmentation completely precise, add a temporary source vertex $\rho$ and source arcs $e_y=(\rho,y^-)$ for $y \in I \cup \{x\}$. The old linkage $\mathcal P$ is an integral unit-capacity flow $f_{\mathcal P}$: it uses $e_i$ for each $i \in I$, then follows the split path corresponding to $P_i$, and finally enters $\omega$ through the terminal arc $q_{t_i}$. Because $x \notin I$, the source arc $e_x$ is unused by $f_{\mathcal P}$. Therefore the arc $e_x$ followed by $Q$ is a residual path from $\rho$ to $\omega$. We augment along this path. For every forward residual arc, we change its flow from $0$ to $1$; for every backward residual arc, we change the flow on the corresponding original arc from $1$ to $0$. These are the only possible changes, so the capacity bounds remain valid: no arc receives flow larger than $1$, and no arc receives negative flow. The conservation check is local. At an internal vertex of the residual path, the augmentation changes exactly one incoming incidence and one outgoing incidence, so the net inflow minus outflow remains unchanged. At $\rho$, one additional unit leaves along $e_x$; at $\omega$, one additional unit arrives. Hence the augmented object is an integral flow of value $|I|+1$ from the source arcs indexed by $I \cup \{x\}$ to $\omega$. An integral unit-capacity flow decomposes into directed paths and directed cycles by repeatedly starting at a source arc with positive flow and following a positive-flow outgoing arc until $\omega$ is reached; finiteness prevents infinite continuation, and any cycle encountered can be removed without changing the source-to-sink value. Since the value is $|I|+1$ and the only available source arcs are the $|I|+1$ arcs $e_y$ with $y \in I \cup \{x\}$, after deleting cycles we obtain one positive-flow path from each $y^-$ to $\omega$. Finally we project the split paths back to the original directed graph $D$. A transit segment $u^+ \to v^-$ becomes the original arc $(u,v) \in A$, and a capacity arc $v^- \to v^+$ records use of the original vertex $v$. Since every capacity arc has capacity one, two projected paths cannot share an original vertex. Since every terminal arc $q_t=(t^+,\omega)$ has capacity one, the terminal endpoints in $T$ are distinct. Thus the projected paths form a linkage from $I \cup \{x\}$ to $T$.[/guided]

[step:Use the augmenting path theorem to force a residual route from $J \setminus I$] Let $\mathcal Q=(Q_j)_{j\in J}$ be a linkage from $J$ to $T$, which exists because $J\in\mathcal I$. Add a temporary source vertex $\rho$ to the split graph and add source arcs $e_y=(\rho,y^-)$ for every $y \in I \cup J$. Give every source arc, capacity arc, transit arc, and terminal arc capacity $1$. The linkage $\mathcal P$ gives an integral flow of value $|I|$ by using the source arcs $e_i$ for $i \in I$. The linkage $\mathcal Q$ gives an integral flow of value $|J|$ in the same network by using the source arcs $e_j$ for $j \in J$. Suppose, for contradiction, that no vertex $x \in J \setminus I$ has a directed residual path from $x^-$ to $\omega$ with respect to the flow determined by $\mathcal P$. The only unused source arcs out of $\rho$ that could start a new augmenting path are the arcs $e_x$ with $x \in J \setminus I$; the arcs $e_i$ with $i \in I$ already carry flow and therefore appear only in the reverse direction in the residual network. Hence there is no residual path from $\rho$ to $\omega$. By the finite integral augmenting-path theorem for unit-capacity networks, a feasible integral flow is maximum exactly when its residual network has no directed path from the source to the sink. The network is finite, all capacities are nonnegative integers, and the residual network is the standard residual network obtained by retaining unused arcs forward and used arcs backward. Therefore the flow determined by $\mathcal P$ would be a maximum flow in this temporary network, so every flow would have value at most $|I|$. This contradicts the flow of value $|J|$ determined by $\mathcal Q$, since $|J|>|I|$. Thus there exists $x \in J \setminus I$ and a residual directed path from $x^-$ to $\omega$. By the augmenting step, $I\cup\{x\}$ is linkable to $T$. [/step]

[step:Conclude that the linkable sets are the independent sets of a matroid] We have shown that $\varnothing \in \mathcal I$, that $\mathcal I$ is closed under taking subsets, and that whenever $I,J \in \mathcal I$ satisfy $|I|<|J|$, there exists $x \in J \setminus I$ such that $I \cup \{x\} \in \mathcal I$. These are exactly the independent-set axioms for a matroid on the finite ground set $V$. Therefore the subsets of $V$ linkable to $T$ are the independent sets of a matroid. [/step]