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# Lecture 15 - Chapter 1: Preserving Joins and Meets

edited February 18

Joins and meets are much more deeply connected to left and right adjoints than I've admitted so far. To dig a little deeper, let's prove this result I mentioned last time:

Theorem. Left adjoints preserve joins and right adjoints preserve meets. Suppose $$f : A \to B$$ and $$g : B \to A$$ are monotone functions between posets. Suppose that $$f$$ is the left adjoint of $$g$$, or equivalently, $$g$$ is the right adjoint of $$f$$. If the join of $$a,a' \in A$$ exists then so does the join of $$f(a), f(a') \in B$$, and

$$f(a \vee a') = f(a) \vee f(a').$$ If the meet of $$b,b' \in B$$ exists then so does the meet of $$g(b), g(b') \in A$$, and

$$g(b \wedge b') = g(b) \wedge g(b').$$ Proof. First, suppose $$f : A \to B$$ is any monotone function and the join $$a \vee a'$$ exists. By definition, $$a \vee a'$$ is the least upper bound of $$a$$ and $$a'$$, so

$$a \le a \vee a' \textrm{ and } a' \le a \vee a' .$$ Since $$f$$ is monotone this gives

$$f(a) \le f(a \vee a') \textrm{ and } f(a') \le f(a \vee a') .$$ This says that $$f(a \vee a')$$ is an upper bound of $$f(a)$$ and $$f(a')$$.

To finish the job, we need to bring in our assumption that $$f$$ is the left adjoint of $$g$$. Let's show that $$f(a \vee a')$$ is the least upper bound of $$f(a)$$ and $$f(a')$$. So, suppose $$b$$ is any other upper bound of these guys:

$$f(a) \le b \textrm{ and } f(a') \le b.$$ Since $$f$$ is the left adjoint of $$g$$, this means

$$a \le g(b) \textrm{ and } a' \le g(b).$$ This says that $$g(b)$$ is an upper bound of $$a$$ and $$a'$$. Since $$a \vee a'$$ is the least upper bound, we have

$$a \vee a' \le g(b) .$$ Since $$f$$ is the left adjoint of $$g$$, this means

$$f(a \vee a') \le b.$$ So, $$f(a \vee a')$$ is indeed the least upper bound of $$f(a)$$ and $$f(a')$$. This implies that $$f(a) \vee f(a')$$ actually exists, and that

$$f(a \vee a') = f(a) \vee f(a').$$ The argument that right adjoints preserve meets works exactly the same way, with all the inequalities reversed. $$\quad \blacksquare$$

To check that you understand this material, come up with the similar argument for right adjoints when you're on a boring train or airplane trip - without looking at any notes.

It's worth noting a little spinoff of the argument I gave. If $$f : A \to B$$ is any monotone function between posets,

$$f(a) \vee f(a') \le f(a \vee a')$$ when the joins here actually exist. Assuming that $$f$$ is a left adjoint gives us the reverse inequality. Similarly, if $$g : B \to A$$ is any monotone function between posets,

$$g(b \wedge b') \le g(b) \wedge g(b')$$ when the meets actually exist. Assuming that $$g$$ is a right adjoint gives us the reverse inequality.

Our next step will be to strengthen the theorem we proved today. The idea doesn't just apply to "binary" meets and joins - that is, meets and joins of a pair of elements. It applies to meets and joins of arbitrary subsets. And when we state it that way, a kind of converse is true too.

Suppose, for example, we have two posets that have all meets and joins. Then a monotone map between them preserves joins of arbitrary subsets if and only if it's a left adjoint! And it preserves meets of arbitrary subsets if and only if it's a right adjoint!

This is staggeringly beautiful. It's called the "Adjoint Functor Theorem for Posets". We'll prove it next time.

Why does it have that name? It's a special case of a more general theorem about adjoint functors for categories. You see, all our results about posets can be generalized to categories, since:

The terms in blue are some of the most important concepts in category theory. Fong and Spivak's strategy is to help you develop an intuition for these concepts by working with posets before diving into more general categories. The Adjoint Functor Theorem, in particular, is rather tricky to prove. But the version for posets is much easier.

To read other lectures go here.

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1.
This says that \$$f(a \vee a') \$$ is an upper bound of \$$f(a) \$$ and \\ f(a') \\).


There's a missing parenthesis in the last part \\ f(a') \\). This is repeated a few lines later.

Comment Source: This says that \$$f(a \vee a') \$$ is an upper bound of \$$f(a) \$$ and \\ f(a') \\). There's a missing parenthesis in the last part \\ f(a') \\). This is repeated a few lines later. 
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2.

Thanks, Jared! Fixed.

Comment Source:Thanks, Jared! Fixed.
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3.
edited April 2018

Are the inequalities reversed in the following statements?

"It's worth noting a little spinoff of the argument I gave. If $$f : A \to B$$ is any monotone function between posets,

$$f(a \vee a') \le f(a) \vee f(a')$$ when the joins here actually exist. Assuming that $$f$$ is a left adjoint gives us the reverse inequality. Similarly, if $$g : B \to A$$ is
any monotone function between posets, $$g(b \wedge g') \ge g(b) \wedge g(b')$$

Because $$a \vee a' \ge a$$ and $$a \vee a' \ge a'$$, therefore $$f(a \vee a') \ge f(a)$$ and $$f(a \vee a') \ge f(a')$$ .

Comment Source:Are the inequalities reversed in the following statements? > "It's worth noting a little spinoff of the argument I gave. If \$$f : A \to B \$$ is _any_ monotone function between posets, > > $$f(a \vee a') \le f(a) \vee f(a')$$ > > when the joins here actually exist. Assuming that \$$f \$$ is a left adjoint gives us the reverse inequality. > Similarly, if \$$g : B \to A \$$ is <i>any</i> monotone function between posets, > > $$g(b \wedge g') \ge g(b) \wedge g(b')$$ > Because \$$a \vee a' \ge a \$$ and \$$a \vee a' \ge a' \$$, therefore \$$f(a \vee a') \ge f(a) \$$ and \$$f(a \vee a') \ge f(a') \$$ .
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4.
edited April 2018

Matias - good catch! I'll fix those. Also, $$g(b∧g′)$$ should be $$g(b∧b′)$$.

Comment Source:Matias - good catch! I'll fix those. Also, \$$g(b∧g′)\$$ should be \$$g(b∧b′)\$$.
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5.

Explaining this idea in terms of posets gives a nice, detailed and intuitive view of this classic fact about adjoints. Thanks for doing these lectures so thoroughly.

Comment Source:Explaining this idea in terms of posets gives a nice, detailed and intuitive view of this classic fact about adjoints. Thanks for doing these lectures so thoroughly.
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6.

Thanks, Christian!

Comment Source:Thanks, Christian!
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7.

I don't understand the final sentence of the proof: why does $$f(a \vee a') = f(a) \vee f(a')$$?

So, $$f(a \vee a')$$ is indeed the least upper bound of $$f(a)$$ and $$f(a')$$. This implies that $$f(a) \vee f(a')$$ actually exists, and that

$$f(a \vee a') = f(a) \vee f(a').$$

Comment Source:I don't understand the final sentence of the proof: why does \$$f(a \vee a') = f(a) \vee f(a') \$$? So, \$$f(a \vee a') \$$ is indeed the _least_ upper bound of \$$f(a) \$$ and \$$f(a') \$$. This implies that \$$f(a) \vee f(a') \$$ actually exists, and that $f(a \vee a') = f(a) \vee f(a').$ 
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8.

@Charles Clingen

(You can quote a piece of text by putting > and a space before each line. It seems to work even for TeX.)

So, $$f(a \vee a')$$ is indeed the least upper bound of $$f(a)$$ and $$f(a')$$. This implies that $$f(a) \vee f(a')$$ actually exists, and that

$$f(a \vee a') = f(a) \vee f(a').$$

I believe that step is using the definition of join, no fancy tricks.

$$f(a) \vee f(a')$$

$$= \textrm{The join is defined as the least upper bound of the two objects.}$$

$$= f(a \vee a') \textrm{, which was shown to be the least upper bound.}$$

Comment Source:@Charles Clingen (You can quote a piece of text by putting > and a space before each line. It seems to work even for TeX.) > So, \$$f(a \vee a') \$$ is indeed the _least_ upper bound of \$$f(a) \$$ and \$$f(a') \$$. This implies that \$$f(a) \vee f(a') \$$ actually exists, and that > $f(a \vee a') = f(a) \vee f(a').$ I believe that step is using the definition of join, no fancy tricks. \$$f(a) \vee f(a')\$$ \$$= \textrm{The join is defined as the least upper bound of the two objects.}\$$ \$$= f(a \vee a') \textrm{, which was shown to be the least upper bound.}\$$
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9.

@ Pete Morcos

Thanks for explaining the last step of the proof. I knew it had to be simple, but I just didn't see it. I have been told more than once by a very wise person that most proofs rely heavily on the use of definitions and I need to make use of that fact. Ah well, practice makes perfect. I really appreciate your clear explanation.

And thank you for explaining a quick way to quote a piece of text too.

You made my day!

Comment Source:@ Pete Morcos Thanks for explaining the last step of the proof. I knew it had to be simple, but I just didn't see it. I have been told more than once by a very wise person that most proofs rely heavily on the use of definitions and I need to make use of that fact. Ah well, practice makes perfect. I really appreciate your clear explanation. And thank you for explaining a quick way to quote a piece of text too. You made my day!