I'm studying category theory by myself and I just came across this sentence from Wikipedia:

An adjunction between categories C and D is somewhat akin to a "weak form" of an equivalence between C and D, and indeed every equivalence is an adjunction. In many situations, an adjunction can be "upgraded" to an equivalence, by a suitable natural modification of the involved categories and functors.

Can someone provide an example of an "upgrade" of an adjunction to an equivalence? I'm interested in understanding why I could intuitively think of an adjunction as a weak form of an equivalence.

  • 1
    $\begingroup$ Abstractly speaking an equivalence is just an adjunction with the unit (or counit) a natural isomorphism; an adjunction is a pair $F,G:\mathcal{C}\rightleftarrows\mathcal{D}$ together with a natural transformation $\eta:1_\mathcal{C}\rightarrow G\circ F$ satisfying a universal property, while an equivalence is all this plus the requirement that $\eta$ be a natural isomorphism (we can prove the existence of the second natural isomorphism $\epsilon:F\circ G\rightarrow 1_\mathcal{D}$ from just the above). For a specific example I will think on it, but this is the essence of the difference. $\endgroup$ – Alec Rhea Dec 1 '18 at 17:49
  • 3
    $\begingroup$ Although the answers below have done an excellent job of trying to interpret the quote as written, I would argue that the quote is actually rather misleading, and if any of these examples are what its author had in mind then it could be better phrased to convey that. $\endgroup$ – Mike Shulman Dec 2 '18 at 2:02
  • 2
    $\begingroup$ The quote is pretty vague about what is meant by "suitable modification" so maybe it's good to advertise two other relevant results. (1) An adjunction induces a homotopy equivalence of classifying spaces, or on the categories obtained by inverting all morphisms on both sides, and (2) the Barr-Beck theorem (en.wikipedia.org/wiki/Beck%27s_monadicity_theorem) gives conditions on when a right adjoint induces an equivalence to the category of algebras over the associated monad on the target $\endgroup$ – Dylan Wilson Dec 2 '18 at 21:54
  • 1
    $\begingroup$ I have made a comment on the Wikipedia talk page of the user who put in this comment (AxelBoldt). $\endgroup$ – Neil Strickland Dec 7 '18 at 19:26
  • 2
    $\begingroup$ Just to set the record straight, when I wrote the above sentence, I had the following examples in mind that were already mentioned below: the Galois correspondence in field theory arising from an adjunction, and numerous equivalences of homotopy categories arising from Quillen adjunctions. $\endgroup$ – Axel Boldt Dec 7 '18 at 23:31

Let $\mathcal C$ and $\mathcal D$ be two categories, and let $F\colon\mathcal C\longrightarrow \mathcal D$ and $G\colon\mathcal D\longrightarrow\mathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $F\circ G\longrightarrow \operatorname{Id}_{\mathcal D}$ and $\operatorname{Id}_{\mathcal C}\longrightarrow G\circ F$ (as mentioned in a comment above).

Let $\mathcal A\subset\mathcal C$ be the full subcategory consisting of all objects $A\in\mathcal C$ for which the natural morphism $A\longrightarrow GF(A)$ is an isomorphism. Similarly, let $\mathcal B\subset\mathcal D$ be the full subcategory consisting of all objects $B\in\mathcal D$ for which the natural morphism $FG(B)\longrightarrow B$ is an isomorphism.

Then one can check that $F(\mathcal A)\subset\mathcal B$ and $G(\mathcal B)\subset\mathcal A$. The restrictions of the adjoint functors $F$ and $G$ to the full subcategories $\mathcal A\subset\mathcal C$ and $\mathcal B\subset\mathcal D$ are again adjoint functors: the functor $F|_{\mathcal A}\colon \mathcal A\longrightarrow\mathcal B$ is left adjoint to the functor $G|_{\mathcal B}\colon \mathcal B\longrightarrow\mathcal A$. The adjunction between the functors $F|_{\mathcal A}$ and $G|_{\mathcal B}$ is an equivalence between the categories $\mathcal A$ and $\mathcal B$, $$ F|_{\mathcal A}\colon\mathcal A\,\simeq\,\mathcal B:\!G|_{\mathcal B}. $$ This result can be found in the paper A. Frankild, P. Jorgensen, "Foxby equivalence, complete modules, and torsion modules", J. Pure Appl. Algebra 174 #2, p.135-147, 2002, https://doi.org/10.1016/S0022-4049(02)00043-9 , Theorem 1.1.

I am not sure whether this should be properly called "upgrading an adjuction to an equivalence", though. The passage from the adjoint pair $(F,G)$ to the equivalence $(F|_{\mathcal A},\,G|_{\mathcal B})$ entails losing rather than gaining information. Perhaps it would be better to call it "restricting an adjunction to an equivalence".

Then again, I do not know what the author of the passage in the Wikipedia article might have had in mind.


Another and probably more natural interpretation of the sentence in the Wikipedia article may be called "localizing an adjunction to an equivalence".

Let $\mathcal C$ and $\mathcal D$ be two categories, and let $F\colon\mathcal C\longrightarrow \mathcal D$ and $G\colon\mathcal D\longrightarrow\mathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $F\circ G\longrightarrow \operatorname{Id}_{\mathcal D}$ and $\operatorname{Id}_{\mathcal C}\longrightarrow G\circ F$, as above.

Denote by $\mathcal S$ the multiplicative class of morphisms in $\mathcal C$ generated by all the morphisms $C\longrightarrow GF(C)$, where $C$ ranges over the objects of $\mathcal C$. Similarly, denote by $\mathcal T$ the multiplicative class of morphisms in $\mathcal D$ generated by all the morphisms $FG(D)\longrightarrow D$, where $D$ ranges over the objects of $\mathcal D$.

Then one can check that the composition $\mathcal C\longrightarrow \mathcal D\longrightarrow \mathcal D[\mathcal T^{-1}]$ of the functor $F$ with the localization functor $\mathcal D\longrightarrow \mathcal D[\mathcal T^{-1}]$ takes all the morphisms from $\mathcal S$ to isomorphisms in $\mathcal D[\mathcal T^{-1}]$. So the functor $F\colon\mathcal C\longrightarrow\mathcal D$ descends to a functor $\overline F\colon\mathcal C[\mathcal S^{-1}]\longrightarrow\mathcal D[\mathcal T^{-1}]$; and similarly the functor $G\colon\mathcal D\longrightarrow\mathcal C$ descends to a functor $\overline G\colon\mathcal D[\mathcal T^{-1}]\longrightarrow\mathcal C[\mathcal S^{-1}]$.

The functors $\overline F$ and $\overline G$ are still adjoint to each other, and this adjunction is an equivalence between the two localized categories: $$ \overline F\colon\mathcal C[\mathcal S^{-1}]\,\simeq\,\mathcal D[\mathcal T^{-1}]:\!\overline G. $$

Yet another and perhaps even more natural interpretation of what may be meant by the sentence in Wikipedia also involves passing to localizations $\mathcal C[\mathcal S^{-1}]$ and $\mathcal D[\mathcal T^{-1}]$ of the given two categories $\mathcal C$ and $\mathcal D$ with respect to some natural multiplicative classes of morphisms (often called the classes of weak equivalences) $\mathcal S\subset\mathcal C$ and $\mathcal T\subset\mathcal D$.

But, rather than hoping that the functors $F$ and $G$ would simply descend to functors between the localized categories, one derives them in some way, producing a left derived functor $$ \mathbb LF\colon\mathcal C[\mathcal S^{-1}]\longrightarrow\mathcal D[\mathcal T^{-1}] $$ and a right derived functor $$ \mathbb RG\colon\mathcal D[\mathcal T^{-1}]\longrightarrow\mathcal C[\mathcal T^{-1}]. $$ Then the functor $\mathbb LF$ is usually left adjoint to the functor $\mathbb RG$, and under certain assumptions they are even adjoint equivalences.

I would not go into further details on derived functors etc. in this answer, but rather suggest some keywords or a key sentence which you could look up: a Quillen equivalence between two model categories induces an equivalence between their homotopy categories.


I think the author of the wikipedia article probably had in mind Leonid Positselski's first answer, where one restricts to the full subcategory of fixed points of the adjunction. Beware there is no guarantee that the fixed points are nonempty! For example, if $F: Set^\to_\leftarrow Ab: U$ is the free/forgetful adjunction bewteen sets and abelian groups, the fixed points are empty.

Here's an illustrative example to have in mind which is not so degenerate. Let $K/k$ be a Galois extension. Then there is an adjuntion between the poset of intermediate subfields $k \subseteq L \subseteq K$ and the opposite of the poset of subgroups of of $Gal(K/k)$; in one direction we send a group to its field of fixed points and in the other direction we send a field to the group of automorphisms that fix it.

This adjunction is typically not an equivalence, but we can pass to the fixed points of the adjunction to get an equivalence between the poset of normal subgroups of $Gal(K/k)$ and the opposite of the poset of intermediate Galois extensions $k \subseteq L \subseteq K$.

Thus the fundamental theorem of Galois theory may be viewed as calculating the fixed point set of an adjunction, and thus as identifying where an adjunction restricts to an equivalence.


I agree that Leonid Positselski’s first answer seems probably what the writer had in mind: given an adjunction, restricting to the categories of “fixed points” on each side yields an equivalence. Here are two important examples in nature, both involving the category of topological spaces:

  • There’s an adjunction between the categories of preordered sets and topological spaces, sending a preordered set $(X,\leq)$ to $X$ with the topology of down-closed sets, and sending a topological space $Y$ to $Y$ with its specialisation order. All preorders are fixpoints; on the other side, the fixpoints are exactly the Alexandrov spaces, i.e. spaces where arbitrary intersections of opens are open. Restricting to this subcategory shows that the category of preorders is equivalent (in fact, isomorphic!) to the category of Alexandrov spaces.

  • The adjunction between the categories of topological spaces and locales, sending a topological space to its frame/locale of opens and sending a locale to its space of points, restricts to the equivalence between spatial locales and sober spaces.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.