Every connected tangent map meeting a leaf factors uniquely through that leaf

The set $U$ is nonempty by hypothesis. If $x\in U$, use [L1] to choose a flat chart $\phi =(u,v):W\to {\mathbb{R}}^k\times {\mathbb{R}}^{n-k}$ around $F(x)$, and choose a connected coordinate neighborhood $C$ of $x$ contained in the open set $F^{-1}(W)$. On $C$, each component of $v\circ F$ has zero differential because $dF(TP)\subseteq \mathcal{D}=\ker dv$, so [L2] makes $v\circ F$ constant. Thus [L5] puts $F(C)$ in the plaque through $F(x)$, which lies in $L$. Hence $C\subseteq U$, and $U$ is open.

If $x\in P\setminus U$, the same [L1]–[L2] argument gives a connected open neighborhood $C$ of $x$ whose image lies in one plaque and hence one leaf. That leaf is not $L$, because it contains $F(x)\mathrm{\notin }L$, so $C\subseteq P\setminus U$. Thus $P\setminus U$ is open. Now $U$ is a nonempty clopen subset of the connected space $P$, so [L4] gives $U=P$ and therefore $F(P)\subseteq L$.

The inclusion $j$ is injective by [L3], so step 2.1 forces a unique set map $\tilde{F}:P\to L$ with $j\circ \tilde{F}=F$. Around each $x\in P$, repeat the flat-chart argument of step 1.1 to obtain a connected neighborhood whose image lies in one plaque. That plaque is a smooth coordinate patch of $L$ by [L3], and in its plaque coordinates $\tilde{F}$ has the same smooth coordinate expression as $F$. Hence $\tilde{F}$ is smooth, and injectivity of $j$ gives uniqueness.

Therefore every connected tangent map that meets a leaf factors uniquely through that leaf.