# Lecture 16 - The Adjoint Functor Theorem for Posets

Today I'll prove the first really profound result in this course: the Adjoint Functor Theorem for Posets. It establishes a deep link between left adjoints and joins, and between right adjoints and meets. This is the climax of Chapter 1: if you survive this lecture, the final one will be a downhill slide!

Last time I showed that left adjoints preserve joins and right adjoints preserve meets - but I only considered "binary" meets and joins: that is, the meets and join of a *pair* of elements. We can do much better.

Remember, given any subset \( S \) of a poset \( A \), we say the **join** of \( S \) is the least upper bound of \( S \), if it exists. We denote this join by \( \bigvee S \), but don't be fooled by the notation into thinking it always exists. Similarly, the **meet** or greatest lower bound of \( S \subseteq A \) is denoted by \( \bigwedge S \) - if it exists.

Now for two theorems. I suggest that you read the statements of both these theorems and ponder them a bit before reading the proofs. The statements are ultimately more important than the proofs, so don't be demoralized if you find the proofs tricky!

**Theorem.** If a monotone function \( f: A \to B \) is a left adjoint, it preserves joins whenever they exist. That is, whenever a set \( S \subseteq A \) has a join we have

[ f (\bigvee S ) = \bigvee \{ f(a) : \; a \in S\} . ]

Similarly, if a monotone function between posets \(g : B \to A \) is a right adjoint, it preserves meets whenever they exist. That is, whenever a set \( S \subseteq B \) has a meet we have

[ g(\bigwedge S) = \bigwedge \{ g(b) : \; b \in S \}. ]

**Proof.** We'll just prove the first half, since the second works the same way. We'll assume \( f: A \to B \) is a left adjoint, meaning that it *has* a right adjoint \( g: B \to A \), and we'll show that \(f\) preserves joins whenever they exist. This is very similar to the proof in Lecture 15, but I'll run through the argument again because it's so important. I'll go a bit faster this time!

Suppose \(S \subseteq A\) has a join \(j = \bigvee S\). This implies that \(a \le j\) for all \( a \in S \), so \(f(a) \le f(j) \), so \( f(j) \) is an upper bound of \( \{ f(a) : \; a \in S\} \). We just need to show it's the *least* upper bound of this set. So, suppose \( b \in B \) is any other upper bound of this set. This means that

[ f(a) \le b ]

for all \(a \in S \), but thanks to the magic of adjoints this gives

[ a \le g(b) ]

for all \(a \in S \) so \( g(b) \) is an upper bound of \( S\). Since \( j \) is the *least* upper bound we conclude

[ j \le g(b) , ]

but thanks to the magic of adjoints this gives

[ f(j) \le b ]

so \( f(j) \) is indeed the *least* upper bound of \( \{ f(a) : \; a \in S\} \). \( \qquad \blacksquare \)

Okay, that was fun. But now comes the really exciting part: a kind of converse is true too! It's easiest to state when we have lots of joins or meets available. We say a poset **has all joins** if every subset has a join, and similarly for meets:

**Adjoint Functor Theorem for Posets.** Suppose \(A\) is a poset that has all joins and \( B \) is any poset. Then a monotone map \(f : A \to B \) is a left adjoint if and only if it preserves all joins.

Similarly, suppose \(B\) is a poset that has all meets and \( A \) is any poset. Then a monotone map \(g : B \to A \) is a right adjoint if and only if it preserves all meets.

**Proof.** Again we'll prove only the first half. So, we'll assume \(A\) is a poset that has all joins, \( B \) is any poset, and \(f : A \to B \) is a monotone map.

The previous theorem assures us that if \( f \) is a left adjoint it preserves all joins, so we only need to prove the converse.

Suppose that \(f\) preserves all joins. To show it's a left adjoint, we construct its right adjoint \( g : B \to A \) using the formula given in Lecture 6:

[ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . ]

Since \(A\) has all joins, \(g(b)\) is well-defined. To see that \(g\) is a monotone, note that if \( b \le b' \) then

[ \{a \in A : \; f(a) \le b \} \subseteq \{a \in A : \; f(a) \le b' \} ]

so

[ g(b) = \bigvee \{a \in A : \; f(a) \le b \} \le \bigvee \{a \in A : \; f(a) \le b' \} = g(b') . ]

See why?

Next we show that \(f\) is the left adjoint of \(g\):

[ f(a_0) \le b_0 \textrm{ if and only if } a_0 \le g(b_0) ]

for all \( a_0 \in A, b_0 \in B \).

To show this, first suppose \( f(a_0) \le b_0 \). Then \( a_0 \) is an element of \( \{ a \in A : f(a) \le b_0 \} \), so

[ a_0 \le \bigvee \{ a \in A : f(a) \le b_0 \} = g(b_0) ]

by the definition of \( g \). So, we have \( a_0 \le g(b_0) \) as desired.

Conversely, suppose \( a_0 \le g(b_0) \). Then \(f(a_0)) \le f(g(b_0)) \), so if we can show \(f(g(b_0)) \le b_0 \) then we'll have \( f(a_0) \le b_0 \). For this, note:

[ f(g(b_0)) = f( \bigvee \{a \in A : \; f(a) \le b_0 \}) = \bigvee \{f(a) \in B : \; f(a) \le b_0 \}) \le b_0 ]

where in the middle step we finally use the fact that \(f\) preserves joins. So, we have \( a_0 \le g(b_0) \) as desired - and we're done! \( \qquad \blacksquare \)

So the connection between joins and left adjoints, or meets and right adjoints, is very strong. Next time, in my last lecture on Chapter 1, I'll explain this a bit more.

**To read other lectures go here.**