Yoneda’s Lemma (part 1)
So I think we’re ready, at least for the statement of Yoneda’s lemma. It says that for any locally small category , if is an object in , and a functor, then Moreover, the isomorphism is natural in both...
View ArticleCoproducts in the category of Sets
As with products. Not every category has coproducts. However, does. I claim that given two sets and , their coproduct is the disjoint union of and . Since coproducts are unique, all we need to do is...
View ArticleMore Pedantry
Similar to last time, I want to talk about something mildly pedantic, but again, very important. There’s something in-between large categories and small categories. We don’t call it a medium category...
View ArticlePedantry
We’ve mentioned the category (the category of sets with functions as morphisms). Consider another category which has two objects and and a morphism between them (along with the required identity...
View ArticleGuest post: Isaac Solomon
Read Isaac’s blog here. From a philosophical point of view, the Axiom of Choice (AC) is often a tough sell. On the one hand, there’s something wrong with vector spaces without bases, or fields without...
View ArticlePhilosophy and choice
Seriously, category theory is coming soon. I promise. In the meantime, a friend (who posts on this awesome blog) and I had a conversation about the axiom of choice from a philosophical standpoint. It...
View ArticleGlobal choice and algebraic closures
It was pointed out to me today that I glazed over a set-theoretic point in my proof that every field has an algebraic closure. We appealed to Zorn’s lemma, which says: Given a partially ordered set ,...
View ArticleEvery field has an algebraic closure
Yeah. That’s right. Every field. Yesterday we showed how we could add a root of a polynomial to a field and that would make it bigger. So we can just keep adding roots over and over again until we get...
View ArticleEvery vector space has a basis
If you haven’t seen a Zorn’s lemma argument before, this will probably be bewildering. It is here for the sake of completeness. Theorem: Every vector space has a basis. Proof: Consider the collection...
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