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.



Is Top_4 (normal spaces) a reflective subcategory of Top_3 (regular spaces)?
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](https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-12/issue-3/Full-reflective-subcategories-and-generalized-covering-spaces/10.1215/ijm/1256054104.full), by Kennison. Kennison lists closure under products as a necessary condition for reflectiveness.


