Lecture 5 - Galois Connections

Okay: I've told you what a Galois connection is. But now it's time to explain why they matter. This will take much longer - and be much more fun.

Galois connections do something really cool: they tell you the best possible way to recover data that can't be recovered.

More precisely, they tell you the best approximation to reversing a computation that can't be reversed.

Someone hands you the output of some computation, and asks you what the input was. Sometimes there's a unique right answer. But sometimes there's more than one answer, or none! That's when your job gets hard. In fact, impossible! But don't let that stop you.

Suppose we have a function between sets, \(f : A \to B\) . We say a function \(g: B \to A \) is the inverse of \(f\) if

[ g(f(a)) = a \textrm{ for all } a \in A \quad \textrm{ and } \quad f(g(b)) = b \textrm{ for all } b \in B ]

Another equivalent way to say this is that

[ f(a) = b \textrm{ if and only if } a = g(b) ]

for all \(a \in A\) and \(b \in B\).

So, the idea is that \(g\) undoes \(f\). For example, if \(A = B = \mathbb{R}\) is the set of real numbers, and \(f\) doubles every number, then \(f\) has an inverse \(g\), which halves every number.

But what if \(A = B = \mathbb{N}\) is the set of natural numbers, and \(f\) doubles every natural number. This function has no inverse!

So, if I say "\(2a = 4\); tell me \(a\)" you can say \(a = 2\). But if I say "\(2a = 3\); tell me \(a\)" you're stuck.

But you can still try to give me a "best approximation" to the nonexistent natural number \(a\) with \(2 a = 3\).

"Best" in what sense? We could look for the number \(a\) that makes \(2a\) as close as possible to 3. There are two equally good options: \(a = 1\) and \(a = 2\). Here we are using the usual distance function, or metric, on \(\mathbb{N}\), which says that the distance between \(x\) and \(y\) is \(|x-y|\).

But we're not talking about distance functions in this class now! We're talking about preorders. Can we define a "best approximation" using just the relation \(\le\) on \(\mathbb{N}\)?

Yes! But we can do it in two ways!

Best approximation from below. Find the largest possible \(a \in \mathbb{N}\) such that \(2a \le 3\). Answer: \(a = 1\).

Best approximation from above. Find the smallest possible \(a \in \mathbb{N}\) such that \(3 \le 2a\). Answer: \(a = 2\).

Okay, now work this out more generally:

Puzzle 14. Find the function \(g : \mathbb{N} \to \mathbb{N}\) such that \(g(b) \) is the largest possible natural number \(a\) with \(2a \le b\).

Puzzle 15. Find the function \(g : \mathbb{N} \to \mathbb{N}\) such that \(g(b)\) is the smallest possible natural number \(a\) with \(b \le 2a\).

Now think about Lecture 4 and the puzzles there! I'll copy them here with notation that better matches what I'm using now:

Puzzle 12. Find a right adjoint for the function \(f: \mathbb{N} \to \mathbb{N}\) that doubles natural numbers: that is, a function \(g : \mathbb{N} \to \mathbb{N}\) with

[ f(a) \le b \textrm{ if and only if } a \le g(b) ]

for all \(a,b \in \mathbb{N}\).

Puzzle 13. Find a left adjoint for the same function \(f\): that is, a function \(g : \mathbb{N} \to \mathbb{N}\) with

[ g(b) \le a \textrm{ if and only if } b \le f(a) ]

Puzzle 16. What's going on here? What's the pattern you see, and why is it working this way?

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