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.

\\[

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.