Using Michael Hong's reasoning in 22, I think that,

\\[
g^\ast f^\ast = \\\\
(\cap\_x \otimes \cap\_y \otimes 1\_{z^\ast}); \\\\
(1\_{x^\ast} \otimes f \otimes 1\_{y^\ast} \otimes g \otimes 1\_{y^\ast}); \\\\
(1\_{x^\ast} \otimes \cup\_y \otimes \cup\_z ).
\\]

However, proving that,

\\[
(fg)^\ast = \\\\
(\cap\_x \otimes 1\_{z^\ast}); \\\\
(1\_{x^\ast} \otimes (g \circ f) \otimes 1\_{z^\ast});\\\\
(1\_{x^\ast} \otimes \cup\_z ) \\\\
\\]
\\[
=\\\\
\\]
\\[
(\cap\_x \otimes \cap\_y \otimes 1\_{z^\ast});\\\\
(1\_{x^\ast} \otimes f \otimes 1\_{y^\ast} \otimes g \otimes 1\_{y^\ast});\\\\
(1\_{x^\ast} \otimes \cup\_y \otimes \cup\_z ) \\\\
= g^* f^*
\\]

Is a bit harder.