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Just For Fun 5

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The basic laws of arithmetic, like \(a \times (b + c) = a\times b + a \times c\), are secretly laws of set theory. But they apply not only to sets, but to many other structures!

Emily Riehl explained this in Eugenia Cheng's recent "Categories for All" session at a meeting of the Mathematical Association of America. Check out her slides:

To hear her discuss this argument, and math in general, listen to her episode of My Favorite Theorem:

Her proof that \(a \times (b + c) = a\times b + a\times c\) involves three facts about the category of sets:

  • It has binary coproducts (the disjoint union \(A + B\) of two sets).

  • It has binary products (the cartesian product \(A \times B\) of two sets).

  • It is Cartesian closed (there's a set \(\mathrm{Fun}(A,B)\) of functions from \(A\) to \(B\)).

The defining property of the coproduct \(A + B\) is that a function from \(A+B\) to any set \(X\) is 'the same as' a function from \(A\) to \(X\) together with a function from \(B\) to \(X\).

The defining property of the product \(A \times B\) is that a function from any set \(X\) to \(A+B\) is the same as a function from \(X\) to \(A\) together with a function from \(X\) to \(B\).

The defining property of \(\mathrm{Fun}(A,B)\) is that a function from any set \(X\) to \(\mathrm{Fun}(A,B)\) is the same as a function from \(A \times X\) to \(B\). This change of viewpoint is called currying, and Riehl uses it twice in her proof.

She calls the defining property of \(+\) 'pairing', and she uses that twice too. I would call this copairing, but she wisely decided that sounds too technical.

Her proof actually never uses the defining property of \(\times\), which I would call pairing.

Her argument works for any category that has the necessary properties! For example, the category of topological spaces and continuous maps. But it's nice to see that lurking inside grade-school arithmetic there is an introduction to category theory.

Comments

  • 1.

    This is great proof but I can't help noticing that the argument doesn't work in Top because it's not cartesian closed. Nevertheless the result holds. Now I'm wondering why!

    Comment Source:This is great proof but I can't help noticing that the argument *doesn't* work in **Top** because it's not cartesian closed. Nevertheless the result holds. Now I'm wondering why!
  • 2.

    Hmm, it was pretty dumb of me to give the example of topological spaces: these days, algebraic topologists usually tweak the category \(\mathbf{Top}\) to get a 'convenient category' of topological spaces that is cartesian closed. I was just trying to grab ahold of some obvious category where \(\times\) distributes over \(+\). I should have taken something like the category of graphs.

    I don't know a beautiful formal explanation for why \(\times\) distributes over \(+\) in \(\mathbf{Top}\).

    Comment Source:Hmm, it was pretty dumb of me to give the example of topological spaces: these days, algebraic topologists usually tweak the category \\(\mathbf{Top}\\) to get a ['convenient category'](https://ncatlab.org/nlab/show/convenient+category+of+topological+spaces) of topological spaces that _is_ cartesian closed. I was just trying to grab ahold of some obvious category where \\(\times\\) distributes over \\(+\\). I should have taken something like the category of graphs. I don't know a beautiful formal explanation for why \\(\times\\) distributes over \\(+\\) in \\(\mathbf{Top}\\).
  • 3.

    Here's something that's been on my mind for the past month, but haven't had the opportunity to ask, can we categorify the zero-product property?

    \[ x \times y = 0 \Leftrightarrow a =0 \lor b =0 \]

    Comment Source:Here's something that's been on my mind for the past month, but haven't had the opportunity to ask, can we categorify the zero-product property? \\[ x \times y = 0 \Leftrightarrow a =0 \lor b =0 \\]
  • 4.

    Also, the zero-property rule also works for equations and inequalities.

    For instance the equation

    \[ 0 = x^2 + y^2 -1 \]

    and

    \[ 0 = |x+y|+|x-y|-2 \]

    can be combined to give,

    \[ 0 = (x^2 + y^2 -1)(|x+y|+|x-y|-2). \]

    Comment Source:Also, the zero-property rule also works for equations and inequalities. For instance the equation \\[ 0 = x^2 + y^2 -1 \\] and \\[ 0 = |x+y|+|x-y|-2 \\] can be combined to give, \\[ 0 = (x^2 + y^2 -1)(|x+y|+|x-y|-2). \\]
  • 5.
    edited August 13

    Interesting question, Keith! I guess we can look for categories where a product \(x \times y\) is initial iff either \(x\) or \(y\) is initial:

    $$ x \times y \cong 0 \quad \iff \quad x \cong 0\textrm{ or } y \cong 0. $$ Examples include the category of sets and the category of topological spaces. Counterexamples include the category of vector spaces and the category of abelian groups.

    Either implication here seems somewhat interesting, because 'initial object' is a colimit and 'product' is a limit, and generally limits and colimits don't play well together without extra conditions. The backwards equation \( \Leftarrow \) is true in any cartesian closed category, I believe, since then \(x \times -\) is a left adjoint, so it preserves colimits.

    The concept of 'strict initial object' seems relevant:

    Comment Source:Interesting question, Keith! I guess we can look for categories where a product \\(x \times y\\) is initial iff either \\(x\\) or \\(y\\) is initial: \[ x \times y \cong 0 \quad \iff \quad x \cong 0\textrm{ or } y \cong 0. \] Examples include the category of sets and the category of topological spaces. Counterexamples include the category of vector spaces and the category of abelian groups. Either implication here seems somewhat interesting, because 'initial object' is a colimit and 'product' is a limit, and generally limits and colimits don't play well together without extra conditions. The backwards equation \\( \Leftarrow \\) is true in any cartesian closed category, I believe, since then \\(x \times -\\) is a left adjoint, so it preserves colimits. The concept of 'strict initial object' seems relevant: * nLab, [Strict initial object](https://ncatlab.org/nlab/show/initial+object#Strict).
  • 6.
    edited August 14

    If we can transform any two functions \(f,g : x \to y\), where \(f(x) =y\) and \(g(x)=y\), into standard form \((y-f(x))=0\) and \((y-g(x))=0\), and then those two standard forms combined via the zero-product property, \((y-f(x)) \times (y-g(x)) = 0\), can we do the same with sets?

    For instance, given sets \(X\) and \(Y\), and two function \(f,g : X \to Y\), can we create a 'standard form' for the function of sets, \((Y \setminus f(X)) \cong \varnothing\) and \((Y \setminus g(X)) \cong \varnothing\), and then can we combine the two 'standard forms' via the zero-Cartesian product property, \((Y \setminus f(X)) \times (Y \setminus g(X)) \cong \varnothing\)?

    Comment Source:If we can transform any two functions \\(f,g : x \to y\\), where \\(f(x) =y\\) and \\(g(x)=y\\), into standard form \\((y-f(x))=0\\) and \\((y-g(x))=0\\), and then those two standard forms combined via the zero-product property, \\((y-f(x)) \times (y-g(x)) = 0\\), can we do the same with sets? For instance, given sets \\(X\\) and \\(Y\\), and two function \\(f,g : X \to Y\\), can we create a 'standard form' for the function of sets, \\((Y \setminus f(X)) \cong \varnothing\\) and \\((Y \setminus g(X)) \cong \varnothing\\), and then can we combine the two 'standard forms' via the zero-Cartesian product property, \\((Y \setminus f(X)) \times (Y \setminus g(X)) \cong \varnothing\\)?
  • 7.
    edited August 14

    Are you using \(f \colon x \to y \) to mean that \(f\) is a function with \(f(x) = y\)? Nobody does that. We write \(f \colon x \mapsto y \) for this purpose.

    The reason is that \(f \colon x \to y\) means that \(f \) is a function from the set \(x\) to the set \(y\). Admittedly it's unconventional to use lower-case letters as names for sets, but if you write \(f \colon x \to y\) and tell me \(f\) is a function, I have to conclude it's a function from the set \(x\) to the set \(y\).

    Anyway, sorry to nitpick, but my brain is melting from strange notation before I even get to the actual question. I think the answer is "no".

    Comment Source:Are you using \\(f \colon x \to y \\) to mean that \\(f\\) is a function with \\(f(x) = y\\)? Nobody does that. We write \\(f \colon x \mapsto y \\) for this purpose. The reason is that \\(f \colon x \to y\\) means that \\(f \\) is a function from the set \\(x\\) to the set \\(y\\). Admittedly it's unconventional to use lower-case letters as names for sets, but if you write \\(f \colon x \to y\\) and tell me \\(f\\) is a function, I have to conclude it's a function from the set \\(x\\) to the set \\(y\\). Anyway, sorry to nitpick, but my brain is melting from strange notation before I even get to the actual question. I think the answer is "no".
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