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I’m studying some category theory by reading Mac Lane linearly and solving exercises.

In question 5.9.4 of the second edition, the reader is asked to construct left adjoints for each of the inclusion functors $\mathbf{Top}_{n+1} $ in $\mathbf{Top}_n$, for $n=0, 1, 2, 3$, where $\mathbf{Top}_n$ is the full subcategory of all $T_n$-spaces in Top, with $T_4$=Normal, $T_3$=Regular, etc.

For $n=0, 1, 2$, it seems to me that I can use the AFT, with the solution set constructed similarly to the one constructed for proving that Haus (=$\mathbf{Top}_2$) is a reflective subcategory of Top (Proposition 5.9.2, p. 135 of Mac Lane).

But I can’t figure out what should I do with the case of $n=3$, that is, with the inclusion functor $\mathbf{Top}_4$ in $\mathbf{Top}_3$: $\mathbf{Top}_4$ doesn’t even have products, so it seems that I cannot use the AFT.

Is there some direct construction of this left adjoint (by universal arrows, perhaps)? Answers including a reference would be especially helpful.

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    $\begingroup$ Hmm. I am able to show the subcategory of completely regular spaces is reflective, but the argument does not extend to normal spaces, afaict. $\endgroup$
    – Mariano Suárez-Álvarez
    Dec 21, 2009 at 23:02
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    $\begingroup$ @GerryMyerson, thank you for correcting my wrong correction. I was so convinced that "Mac Lane" was a typo that I couldn't even bring myself to check it. (But, I say consolingly to myself, at least I spell "Macdonald" without an intercap.) $\endgroup$
    – LSpice
    Jul 16, 2022 at 5:49
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    $\begingroup$ @LSp, searching this website for "MacLane" yielded over 600 hits, so you have a lot of company. $\endgroup$
    – Gerry Myerson
    Jul 16, 2022 at 9:22
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    $\begingroup$ @LSpice According to Mac Lane's autobiography, his spelling of his surname originated as a typo by his wife (which he decided to keep). $\endgroup$
    – Robert Furber
    Jul 16, 2022 at 16:30
  • $\begingroup$ @RobertFurber, ah, thanks! So I was inadvertently right that Mac Lane was a typo, but @‍user2734 and @‍GerryMyerson were more right that it was the correct spelling. Interesting! $\endgroup$
    – LSpice
    Jul 16, 2022 at 17:13

1 Answer 1

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I think that MacLane made a mistake. I think that he just forgot that the category of $T_4$ spaces lacks closure properties.

Claim: If $\mathcal{A} \subseteq \mathcal{C}$ is a (full) reflective subcategory and $\mathcal{C}$ has finite products, then $\mathcal{A}$ is closed under $\mathcal{C}$'s finite products, up to isomorphism.

If $\mathcal{A}$ is reflective, it means that an object $X \in \mathcal{C}$ has an "$\mathcal{A}$-ification" $X'$, e.g., an abelianization in the case where $\mathcal{A}$ is abelian groups and $\mathcal{C}$ is groups. If $A, B \in \mathcal{A}$ are objects, then they have a product $A \times B$ in $\mathcal{C}$, and then that has an $\mathcal{A}$-ification $(A \times B)'$. There is an $\mathcal{A}$-ification morphism $A \times B \to (A \times B)'$, and there are also projection morphisms $A \times B \to A, B$. Since $A, B \in \mathcal{A}$, the projection morphisms factor through $(A \times B)'$, and then the universal property gives you a morphism $(A \times B)' \to A \times B$. So you get canonical morphisms in both directions between $A \times B$ and $(A \times B)'$, and I think that some routine bookkeeping shows that they are inverses.

So the Sorgenfrey plane (which is the non-normal Cartesian square of the Sorgenfrey line) does not have a "$T_4$-ification".

I got onto this track after I found the paper, Reflective subcategories and generalized covering spaces, by Kennison. Kennison lists closure under products as a necessary condition for reflectiveness.

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    $\begingroup$ Greg's claim also follows from a well-known categorical fact: that if A is a reflective subcategory of a category C then the inclusion of A into C is monadic. (For example, the inclusion of abelian groups into groups is monadic.) Monadic functors create limits, and in particular finite products, giving Greg's Claim. $\endgroup$
    – Tom Leinster
    Dec 22, 2009 at 12:36
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    $\begingroup$ @Tom: I suppose that monadic functors capture and generalizes the ideas in the proof that I sketch? @unknown: You're welcome. If I may make a request too, can you register with your name? $\endgroup$
    – Greg Kuperberg
    Dec 22, 2009 at 17:46
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    $\begingroup$ Greg: yes, the general proof that monadic functors create limits, when specialized to this case, is indeed the same as your proof. (I don't think there could be two substantially different proofs of your claim.) One might say that the inclusion of abelian groups into groups is monadic "because" the subcategory is obtained by imposing an equation in the sense of universal algebra, and monadic functors are all about universal algebra. For instance, every forgetful functor mentioned in part 1 of my answer to mathoverflow.net/questions/5786 is monadic. $\endgroup$
    – Tom Leinster
    Dec 22, 2009 at 19:50

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