2.5.1

Let $$ \phi : S \rightarrow T $$ be a map, and for a given t in T, let $$ \phi^{-1}(t) = \{s \in S: \phi(s)=t\}$$ be the fibre over t. Let $$ F = \{\phi^{-1}(t): t \in T, \phi^{-1}(t) \not= \emptyset \} $$ be the set of all nonempty fibres of $$\phi $$, then we claim F forms a partition of S.

Fix s in S arbitrarily, then $$ \phi(s) = t $$ for some t in T, so s is in $$ \phi^{-1}(t) $$ (which is thus nonempty) and hence s is in $$ \bigcup_{\phi^{-1}(t) \in F} \phi^{-1}(t) $$, showing that S is contained in the union of the nonempty fibres. On the other hand, let s be in the union of the nonempty fibres, then s is in some $$\phi^{-1}(t) $$ in F, so that s is in S, by the definition of the fibres. Thus S is equal to the union of the nonempty fibres.

Now we show that distinct members of F are disjoint by proving the contrapositive. Let $$\phi^{-1}(t), \phi^{-1}(t') $$ be members of F with nonempty intersection, then there exists some s in S such that $$\phi(s) = t $$ and also $$\phi(s) = t' $$, so that t = t' and hence $$\phi^{-1}(t) = \phi^{-1}(t') $$. Therefore, distinct members of F are disjoint, so that F is a partition of S.