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Some time ago, in connection with trying to understand a construction of Amitsur (Embeddings of matrix rings, Pac JM 36 (1971), I stumbled across a short paper which considered some variant of the following question:

given (forgetful) functors $H: {\mathcal C} \to {\mathcal B}$ and $G:{\mathcal B}\to {\mathcal A}$, such that $GH:{\mathcal C}\to {\mathcal A}$ and $G:{\mathcal B}\to {\mathcal A}$ both have left adjoints, when can we deduce that $H$ has a left adjoint?

I think in the intended application ${\mathcal A}$ was the category of sets and ${\mathcal B}$ was the category of rings; the idea being that it is easier or more transparent to construct "the free widget on a generating set" than "the free widget on a generating ring".

Unfortunately, I can't remember the name of the author nor the title of the paper. Does this sound familiar to anyone? and if not, does anyone have an alternative reference? Sorry to ask such a vague question, but I have been racking my brains and searching on MathSciNet to no avail.

UPDATE: while reading Todd Trimble's answer below, the veil inexplicably lifted and I was able to remember the author's surname, from which a quick Google turned up what I was after:

An Interpolation Theorem for Adjoint Functors S. A. Huq, Proceedings of the American Mathematical Society Vol. 25, No. 4 (Aug., 1970), pp. 880-883

Sorry to waste people's time!

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In a paper currently in progress, we consider something similar in a step in a larger proof, but we have more to work with. Our category $\mathcal B$ is something like "sheaves" and $\mathcal A$ something like "presheaves", and $G$ something like "$\text{Forget}$", so that its left adjoint "$\text{Sheafify}$" satisfies $\text{Sheafify}\circ\text{Forget}=\text{id}$. (It suffices for $G$ to be full to have its adjoint be a one-sided inverse.) In this case, $H$ has a left adjoint, given by $(\text{adjoint to }GH)\circ G$. Anyway, you weren't reading our paper, as it's not finished yet. –  Theo Johnson-Freyd Jan 5 '11 at 3:33
I forgot to accept an answer. Well, I would really like to accept both, but since I can only accept one I'm going to accept Steve Lack's for the link to Power's paper (it's not entirely clear to me that the examples I was considering are of the form that Todd Trimble's answer discusses) –  Yemon Choi Jan 23 '11 at 2:01
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2 Answers

up vote 5 down vote accepted

This should probably be a comment but I don't know how to include links in comments.

Results of this type are often called adjoint triangle theorems. There are many such: see John Power's paper A unified approach to the lifting of adjoints for a summary (and a unified approach).

A sufficient condition for $H$ to admit a left adjoint is that $C$ has coequalizers and that $G$ is ``of descent type''. (This means that if we form the monad $S$ on $A$ induced by $G$ and its left adjoint, then the induced functor $B\to A^S$ is fully faithful.) This includes the important examples of the type considered in Todd's answer.

for a summary.

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Thanks for the information regarding terminology, and for the link. –  Yemon Choi Jan 6 '11 at 1:37
Hi Steve, links in comments are easy: just provide the full URL (incl. http etc). The whole URL won't appear as link text, only a fraction, but the link will work. This is your link in a comment: archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1988__29_1/… –  David Roberts Jan 6 '11 at 1:40
Thanks for this, Steve. –  Todd Trimble Jan 6 '11 at 2:11
thanks David. –  Steve Lack Jan 6 '11 at 3:55
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"Relatively free" functors have been considered at least since the time of Lawvere's thesis; see for example page 111 of 122 for the case involving finitary algebraic theories.

I don't know where this is written down, but the following construction is pretty general and might suit your purposes. Let $\theta: S \to T$ be a morphism of monads on a category $C$, and suppose that the category of algebras $C^T$ has coequalizers (as happens if for example $C= Set$). Then the forgetful functor

$$C^T \to C^S,$$

which takes a $T$-algebra $(d, \alpha: Td \to d)$ to the $S$-algebra

$$Sd \stackrel{\theta d}{\to} Td \stackrel{\alpha}{\to} d,$$

has a left adjoint. This gives the factorization in the evident case where $\mathcal{A} = C$, $\mathcal{B} = C^S$, and $\mathcal{C} = C^T$.

The left adjoint takes an $S$-algebra $(c, \xi: Sc \to c)$ to the coequalizer of the pair of arrows in $C^T$:

$$T\xi: TSc \to Tc, \qquad TSc \stackrel{T\theta c}{\to} TTC \stackrel{mc}{\to} Tc,$$

where $m: TT \to T$ is the multiplication on $T$. This coequalizer might be written $T \circ_S c$, by analogy with the coequalizer $B \otimes_A M$ where $B$ is an algebra over $A$ and $M$ is an $A$-module.

That this is the left adjoint is a slightly lengthy diagram chase which I can produce if desired. (I think I may have written this up in the nLab somewhere, but just now I can't access the nLab. It might be at a page on algebras over a monad, or on free algebras. If it's not there, I should record the detailed proof at the Lab when I get a chance.)

Edit: I wrote up something quickly here at the nLab, which has a detailed (somewhat pedestrian) proof of the adjunction stated above.

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Thanks Todd - I'll have a look at this when I get back to the office tomorrow morning. As it happens, I started reading your answer and suddenly - and and quite inexplicably - remembered the surname of the author! (The name is S. Huq, but I was misremembering his/her name as Hu, and hence my searches were in vain) –  Yemon Choi Jan 5 '11 at 7:01
I'll take credit for jogging your memory! :-) Yes, that name sounds like a likely source for this type of thing. –  Todd Trimble Jan 5 '11 at 7:04
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