be pairwise distinct planes through $U$ contained in $T_1$. If $\Pi_1\cap\Pi_2\cap\Pi_3$ is a line, then $f(\Pi_1)\cap f(\Pi_2)\cap f(\Pi_3)$ is a line.

Note first that $T_1\cap T_2 = \{U\}$ and $U=B_1\cap B_2$.

(i): For the plane $\Pi$ observe that $Z\in\Sub_{k-2}(V)$ and $Z\subset U$. We have $$\begin{gathered}
    X := \{ \starof(p)\cap T_2\colon p\in\peki_k(V),\ U\in p\subset \Pi\} = \\
      \{ \starof(p)\cap T_2\colon p = \penc(H, B_1),\ H\in[Z, U]_{k-1} \} =
        \{ \penc(H,B_2)\colon H\in[Z, U]_{k-1} \}
  \end{gathered}$$ as $H\subset U\subset B_2$. So, $X$ is the set of lines through $U$ on the plane $[Z, B_2]_k$ contained in $T_2$, which completes this part of the proof.

(ii): Let $\Pi_i = [Z_i, B_1]_k$, where $Z_i\in\Sub_{k-2}(V)$ and $Z_i\subset U$ for $i=1,2,3$. By assumption $H := Z_1+Z_2+Z_3\in\Sub_{k-1}(V)$ and $$\Pi_1\cap\Pi_2\cap\Pi_3 = \penc(H, B_1).$$ Since $f(\Pi_i) = [Z_i, B_2]_k$ for $i=1,2,3$ and $H\subset U\subset B_2$, we get that $$f(\Pi_1)\cap f(\Pi_2)\cap f(\Pi_3) = \penc(H, B_2)$$ and we are through.

Dually to \[lem:top-star-top\] we have the following for stars.

Let $U$ be a point and $S_i = [H_i]_k$, $i=1,2$ be distinct stars of $\M$ with $U\in S_1\cap S_2$.

If $\Pi = [H_1, Y]_k$ is a plane through $U$ contained in $S_1$, then $$\{ \topof(p)\cap S_2\colon p\in\peki_k(V),\ U\in p\subset \Pi\}$$ is the set 