Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 7

A poset $P$ becomes a category with the elements of $P$ as objects and exactly one morphism $x\to y$ when $x\le y$, with no morphism otherwise. Reflexivity supplies the identity morphisms and transitivity supplies composition in a category; uniqueness of morphisms proves associativity. A functor $f:P\to Q$ is exactly a monotone function: it must send $x\le y$ to $f(x)\le f(y)$, and this condition suffices because all the required morphisms are unique. Antisymmetry says that isomorphic objects of this category are equal.

Each hom-set in a poset is either empty or a singleton. Thus a natural bijection $Y(fx,y)\cong X(x,gy)$ is precisely the conditionThe naturality squares commute automatically, because maps between the relevant singleton hom-sets are unique. This is a Galois connection. Equivalently, the adjunction unit and adjunction counit are the inequalities $x\le gf(x)$ and $fg(y)\le y$. These inequalities imply the displayed equivalence: $f(x)\le y$ gives $x\le gf(x)\le g(y)$, and $x\le g(y)$ gives $f(x)\le fg(y)\le y$.

The two adjoints of inverse image on power sets areThese maps and $p^{\ast }(V)=p^{-1}(V)$ are monotone functions on the inclusion-ordered power sets. For $U\subseteq A$ and $V\subseteq B$,The first equivalence follows by following each element of $U$ through $p$. For the second, every fiber over $b\in V$ must lie in $U$, which is precisely the defining condition for $b\in {\forall }_p(U)$. These are the two Galois connections, proving ${\exists }_p⊣p^{\ast }⊣{\forall }_p$. In particular, a point outside $p(A)$ has an empty fiber and belongs to ${\forall }_p(U)$ even when $U$ is empty; omitting this case would give an incorrect right adjoint.

For a presheaf of sets on a topological space $P$, define $LP=P(\varnothing )$ and $RP=P(S)$. Then the adjoints to constant presheaves on open sets areTo prove the first adjunction, a natural transformation $a:P\to \Delta A$ must satisfy $a_U=a_{\varnothing }\circ {\mathrm{res}}_{U,\varnothing }$ by naturality at the inclusion $\varnothing \subseteq U$. Hence it is uniquely determined by a function $P(\varnothing )\to A$. Every such function defines a natural transformation by this formula, because restriction maps compose. This gives $\mathrm{Nat}(P,\Delta A)\cong \mathbf{Set}(P(\varnothing ),A)$.

For the second adjunction, a natural transformation $b:\Delta A\to P$ satisfies $b_U={\mathrm{res}}_{S,U}\circ b_S$, so it is uniquely determined by a function $A\to P(S)$. Every such function defines the remaining components by restriction, giving $\mathrm{Nat}(\Delta A,P)\cong \mathbf{Set}(A,P(S))$. Evaluation acts on presheaf morphisms componentwise, so these object assignments are indeed functors. The use of $P(\varnothing )$ is essential: a constant presheaf of sets has value $A$ there, whereas a sheaf has a singleton value there. No sheaf condition is imposed in this question.