Okay, Charlie: here's what I _should_ have said, and now do:

To prove that sometimes

\[ f^{\ast}(P \vee Q) \ne f^{\ast}(P) \vee f^{\ast}(Q) , \]

we just need one example. So, take \\(P\\) and \\(Q\\) to be these two partitions:



They are partitions of the set

\[ Y = \\{11, 12, 13, 21, 22, 23 \\}. \]

Take \\(X = \\{11,22\\} \\) and let \\(i : X \to Y \\) be the inclusion of \\(X\\) into \\(Y\\), meaning that

\[ i(11) = 11, \quad i(22) = 22 . \]

Then compute everything! \\(11\\) and \\(22\\) are in different parts of \\(i^{\ast}(P)\\):

\[ i^{\ast}(P) = \\{ \\{11\\}, \\{22\\} \\} . \]

They're also in different parts of \\(i^{\ast}(Q)\\):

\[ i^{\ast}(Q) = \\{ \\{11\\}, \\{22\\} \\} .\]

Thus, we have

\[ i^{\ast}(P) \vee i^{\ast}(Q) = \\{ \\{11\\}, \\{22\\} \\} . \]

On the other hand, the join \\(P \vee Q \\) has just two parts:

\[ P \vee Q = \\{\\{11,12,13,22,23\\},\\{21\\}\\} . \]

If you don't see why, figure out the finest partition that's coarser than \\(P\\) and \\(Q\\) - that's \\(P \vee Q \\). Since \\(11\\) and \\(22\\) are in the same parts here, the pullback \\(i^{\ast} (P \vee Q) \\) has just one part:

\[ i^{\ast}(P \vee Q) = \\{ \\{11, 22 \\} \\} . \]

So, we have

\[ i^{\ast}(P \vee Q) \ne i^{\ast}(P) \vee i^{\ast}(Q) \]

as desired.