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Suppose that a category $A$ is reflective in a category $B$ and that the inclusion functor $K:A\to B$ has a left adjoint $F:B\to A.$ Now what does it technically mean that this bijection of sets $$A(Rb,a)\cong B(b,a)$$ is natural in $a$ and $b$? I.e. what squares are supposed to commute? Here $R=KF:B\to A.$ See the MacLanes' CWM page 91 here. I'm also confused with the apperance of $R$ in that natural bijection. I would rather expect $$A(Fb,a)=B(b,Ka).$$

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  • $\begingroup$ @MaliceVidrine I cannot figure out what should commute. I have always difficulties with this kind of exercises. $\endgroup$ – user122424 Aug 23 '18 at 16:53
  • $\begingroup$ Sorry, I misread part of your question (haven't had coffee yet) and I think I understand where you're coming from better. Mac Lane is being careless about notation here in a way that a lot of authors tend to be about reflective subcategories, so that probably doesn't help. I'll try to cook up an answer if no one beats me to it. $\endgroup$ – Malice Vidrine Aug 23 '18 at 17:13
  • $\begingroup$ @MaliceVidrine Tell me what did you misread and what is written? I'm awaiting your answer... $\endgroup$ – user122424 Aug 23 '18 at 17:17
  • $\begingroup$ I misread your first question as asking whether it was natural in $a,b$, not what it meant. The latter question is more understandable, and highlighted some other problems with Mac Lane's presentation. $\endgroup$ – Malice Vidrine Aug 23 '18 at 17:21
  • $\begingroup$ @MaliceVidrine MacLane's presentation is the best textbook on general categories up to date, despite its age.Therefore it deserves one's attention to every detail. $\endgroup$ – user122424 Aug 23 '18 at 17:26
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The way Mac Lane (and an unfortunate number of authors) write this is unhelpful; consequently, a lot of this answer will be about how to avoid being misled by this notational convention. Note that he states that the isomorphism $$A(Rb,a)\cong B(b,a)$$ is natural in $b\in B$ and in $a\in A$. So this points out one way in which the notation is misleading; in the right hand side, if we're expecting the $a$ place to be natural with respect to morphisms in $A$, then really what we're looking at is $B(b,Ka)$. On objects, this is a distinction that doesn't matter, but when we start looking at the naturality condition, it matters quite a lot.

Similarly, the left hand side shows an object that lives in $B$ (namely $Rb$) in a hom-set $A(-,-)$, so what we're actually looking at here is $A(Fb,a)$. The reason is that an inclusion functor like $K$ is always faithful, so that for any $Rf:Rb\to Rb'$, there is a unique $Ff:Fb\to Fb'$ in $A$ which is carried to it under $K$.

Making these substitutions gives you the form you were expecting. What's happened here is that, since subcategories are usually taken to consist of subclasses of objects and arrows, $Ka$ and $a$ will be the same object, and some authors will consequently denote both with $a$; similarly, they don't always distinguish between $Rb$ and $Fb$ since they will be the same object thought of in two different categories. The reason it's helpful to notationally make these distinctions instead of ignoring them can be illustrated once we answer your other question about naturality.

Looking instead at the nice form $A(Fb,a)\cong B(b,Ka)$, and calling the left-to-right half of the isomorphism $\phi$, naturality in both variables means that for $f:a\to a'$ in $A$, $$B(b,Kf)\circ\phi_{b,a}=\phi_{b,a'}\circ A(Fb,f)$$ and for $g:b'\to b$ in $B$ $$B(g,Ka)\circ\phi_{b,a}=\phi_{b',a}\circ A(Fg,a).$$ And similar equalities hold for $\phi^{-1}$.

Note that if we just denoted the right hand side of the isomorphism $B(b,a)$, then it looks like for any $f:a\to a'$ in $B$ we need $B(b,f)$ to behave nicely in our naturality square. But there may be morphisms in $B$ for which there's no counterpart in $A$, and explicitly showing the inclusion functor avoids the temptation to think that any $B$-morphism between objects from $A$ are part of the naturality condition. Likewise, seeing $A(Rb,a)$ should look wrong since $R$ gives you an object in $B$, but $B(Rb,a)$ is actually not the hom-set you want to look at--there may be morphisms $Rb\to a$ in $B$ (by which we really mean $Rb\to Ka$) that are not morphisms in $A$, but it's only these latter that correspond to morphisms $b\to Ka$ in $B$.

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