In trying to think of an intuitive answer to a question on adjoints, I realised that I didn't have a nice conceptual understanding of what an adjoint pair actually is.

I know the definition (several of them), I've read the nlab page (and any good answers will be added there), I've worked with them, I've found examples of functors with and without adjoints, but I couldn't explain what an adjunction is to a five-year-old, the man on the Clapham omnibus, or even an advanced undergraduate.

So how should I intuitively think of adjunctions?

For more background: I'm a topologist by trade who's been learning category theory recently (and, for the most part, enjoying it) but haven't truly internalised it yet. I'm fully convinced of the value of adjunctions, but haven't the same intuition into them as I do for, say, the uniqueness of ordinary cohomology.


8 Answers 8


The example I would give a five-year old is the following.

Take the category $\mathbb R$ whose objects are the real numbers (or perhaps rational numbers for the five-year old) and a single morphism $x \to y$ whenever $x \leq y$. Let $\mathbb Z$ be the full subcategory consisting of the integers. The inclusion $i : \mathbb Z \to \mathbb R$ has a right and a left adjoint: the former is the floor function, the other is the ceiling function.

I think the five-year old will agree that these are approximations so I would then say that left and right adjoints are just jazzed up versions of approximations.

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    $\begingroup$ OOPS - I forgot to post my name - Paul Smith $\endgroup$
    – user1421
    Jan 10, 2011 at 15:10
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    $\begingroup$ That's going to be one amazing five-year-old. $\endgroup$
    – KConrad
    Jan 10, 2011 at 19:55
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    $\begingroup$ Actually, it's the other way around: the left adjoint is the ceiling and the right adjoint is the floor. If $r \in \mathbb{R}$ and $n \in \mathbb{Z}$ then $\lceil r \rceil \le n$ if and only if $r \le n$, and dually $n \le \lfloor r \rfloor$ if and only if $n \le r$. $\endgroup$ Apr 7, 2014 at 6:40
  • $\begingroup$ @QiaochuYuan could you expand why it's the other way around, if so, then suggest an edit to the answer? $\endgroup$ Feb 17, 2016 at 7:53
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    $\begingroup$ The right adjoint is conservative, safe, it does not want to over estimate, hence the floor [rounding down]. The left adjoint is liberal, risky, it wants to make sure it gets credit for partial work, hence the ceil [rounding up]. The bit that is missing is that the categories are both total orders and it is that fact that drives the interpretation. $\endgroup$
    – phreed
    Apr 12, 2018 at 14:23

I like Wikipedia's motivation for an adjoint functor as a formulaic solution to an optimization problem (though I'm biased, because I helped write it). In short, "adjoint" means most efficient and "functor" means formulaic solution.

Here's a digest version of the discussion to make this more precise:

An adjoint functor is a way of giving the most efficient solution to some optimization problem via a method which is formulaic ... For example, in ring theory, the most efficient way to turn a rng (like a ring with no identity) into a ring is to adjoin an element '1' to the rng, adjoin no unnecessary extra elements (we will need to have r+1 for each r in the ring, clearly), and impose no relations in the newly formed ring that are not forced by axioms. Moreover, this construction is formulaic in the sense that it works in essentially the same way for any rng.

The intuitive description of this construction as "most efficient" means "satisfies a universal property" (in this case an initial property), and that it is intuitively "formulaic" corresponds to it being functorial, making it an "adjoint" "functor".

In this asymmetrc interpretation, the theorem (if you define adjoints via universal morphisms) that adjoint functors occur in pairs has the following intuitive meaning:

"The notion that F is the most efficient solution to the (optimization) problem posed by G is, in a certain rigorous sense, equivalent to the notion that G poses the most difficult problem which F solves."

Edit: I like the comment below emphasizing that an adjoint functor is a globally defined solution. If
$G:C\to D$, it may be true that terminal morphisms exist to some $C$'s but not all of them; when they always exist, this guarantees that they extend to define a unique functor $F:D\to C$ such that $F \dashv G$. This result could have the intuitive interpretation "globally defined solutions are always formulaic".

Compare this for example to the basic theorem in algebraic geometry that a global section (of the structure sheaf) of $\mathrm{Spec} (A)$ is always defined by a single element of $A$; the global sections functor is an adjoint functor representable by the formula $Hom(-,\mathrm{Spec}( \mathbb{Z}))$, so this is actually directly related.

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    $\begingroup$ Just to emphasize on part of this response, note that an adjoint pair provides a global solution (category-wide) as opposed to a local solution provided by universal maps. So universality is more general, in that if the universal map exists for every object in the category, the family of universal maps can be used to define an adjoint. $\endgroup$
    – marc
    Nov 23, 2009 at 19:34
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    $\begingroup$ In the last section, should it be $\mathrm{Spec}(\mathbb{Z}[t])$ instead? $\endgroup$
    – QcH
    Jan 2, 2011 at 3:49
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    $\begingroup$ In which sense a functor poses a question? For example, what is the problem posed by the functor $M\rightarrow M\otimes_A B$? $\endgroup$
    – safak
    Mar 19, 2018 at 22:33

Suppose that $F\colon C\to D$ is a functor. Then there are many situations in which thinking of finding left and right adjoints to $F$ as solving approximation problems is very good intuition. So these would constitute functorial ways to approximate objects in $D$ relative to the image of $F$ by objects in $C$ either on the right or the left. I'm not sure I've really managed to word this in a way that conveys what I have in my head but here are some examples (which I have picked because they have a particularly 'approximationy' flavour but I do think that this works reasonably well in general anyway, I think my selection bias here is more skewed toward what I think about regularly).

Torsion theories are very good examples of this principal. For instance the notion of localization with respect to a homology theory in the homotopy category of spectra or more generally the approximating triangles coming from the acyclization and localization functors of a semi-orthogonal decomposition of a triangulated category. Another nice example along these lines is say the standard t-structure on the derived category of modules over some ring. Here we again have two pairs of adjoints and we can think of one as a right approximation by a bounded below complex coming from the unit and the other as a left approximation by a bounded above complex via the counit.

One can also view resolutions in the derived category in this way. For instance we have a right adjoint to the canonical map from $K(Inj R) \to D(R)$ for a ring $R$ where the first category is the homotopy category of complexes of injective $R$-modules which is taking K-injective resolutions. Similarly other sorts of resolutions, envelopes, and covers can be realised by adjunctions.

All of these examples are particularly nice in the sense that we get triangles or short exact sequences describing the object we start with in terms of our complementary approximations (by complementary I mean that there is orthogonality floating around in all of these examples so we have in some sense decomposed our category).

I think things like the adjoint functor theorem and Brown reprensentability become very reasonable from this point of view. One can loosely interpret them as saying provided things are "small enough" to be manageable and one has enough limits/colimits then one can build universal approximations (i.e. adjunctions) by taking coarse approximations and refining them.

I think this philosophy works well with the one given on the wiki page that Andrew Critch linked to.

  • $\begingroup$ This intuition is the intuition behind Freyd's adjoint functor theorem. $\endgroup$ Jan 30, 2013 at 16:07

For a "man on the Clapham omnibus" gloss on it, I think you could do worse than the Stanford Encyclopedia of Philosophy's entry for Category Theory. It describes adjoints as "conceptual inverses", and elaborates on how to see them that way in some of the standard examples.

I guess this is most probably a lower level than you were really asking for. But I think it articulates pretty well one of the less immediately obvious core points of the intuition (at least, my intuition) of what an adjunction is.

Putting this more precisely/abstractly: when we think of generalising isomorphism between objects of a 1-category to something between objects in a 2-category, we might usually think first of isomorphism and equivalence, but adjunction is also such a generalisation.


Adjoints can be viewed intuitively as expressing conceptual duality between mathematical notions.

Examples along these lines include $${\sf Free}\dashv {\sf Forgetful},$$ $$\varinjlim\dashv\Delta\dashv\varprojlim,$$ where $\Delta:\mathcal{C}\to\mathcal{C}^\mathcal{D}$ is the diagonal functor on a (co-)complete category $\mathcal{C}$, $$\exists\dashv*\dashv\forall,$$ where $*:{\sf Form}(\overline x)\to{\sf Form}(\overline x,y)$ is the functor adjoining a variable $y$ not appearing in $\overline x=(x_0\dots,x_n)$ which we quantify existensially or universally over, $$\lceil x\rceil\dashv i\dashv\lfloor{x}\rfloor$$ as noted by user1421 in his answer, where $i:\mathbb{Z}\to\mathbb{R}$ is the standard inclusion - there are many more that belong on this list.

In essence, if two notions are related to eachother closely enough that giving a definition of one notion (in the presence of sufficient ambient structure) defines the other notion completely, they can be expressed as a (possibly $\infty$-) adjoint pair of functors. Conversely, when we see that two notions can be expressed as an adjoint pair of functors it means we can think about one by thinking about the other plus some canonical additional structure.

In the above cases, one of the two notions is significantly simpler than the other -- forgetful functors are simpler than free object functors, diagonal functors are simpler than limit and colimit functors, adjoining a variable is simpler than existentially or universally quantifying, etc. This means we can understand complex notions by understanding simple ones and making them functorial, then considering their left or right adjoint. Although there may be another way to construct whatever notion we end up with, any two right (or left) adjoints to the same functor will be naturally isomorphic, so we've gotten at the definition safely and using tools that are very well understood and easy to manipulate.

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    $\begingroup$ I don't really see how the e.g. hom functor is "the tensor functor plus some canonical additional structure", or vice versa. Do you have some intuition on this example? $\endgroup$
    – D.R.
    Dec 19, 2022 at 7:09
  • $\begingroup$ @D.R. This would be an instance of inspecting one of the appropriate adjoint functor theorem proofs and subbing in the tensor functor in place of the arbitrary left adjoint, then examining how it’s used to define the right adjoint. I’ll edit with details some time in the next few weeks. $\endgroup$
    – Alec Rhea
    Dec 19, 2022 at 21:50

Another intuitive notion for adjoint functors comes from the string diagram notation for $2$-categories. Functors are $1$-morphisms in the $2$-category $Cat$. Functors $D\xleftarrow L C$ and $C\xleftarrow R D$ are adjoint $L\dashv R$ iff there are natural transformations (i.e. 2-morphisms) $R\circ L\xleftarrow\eta 1_C$ and $1_D\xleftarrow\epsilon L\circ R$ such that the composites $$L\xleftarrow{\epsilon\circ 1_L}L\circ R\circ L\xleftarrow{1_L\circ\eta}L$$ $$R\xleftarrow{1_R\circ\epsilon}R\circ L\circ R\xleftarrow{\eta\circ 1_R}R$$ are equal to the identities $1_L,1_R$ respectively.

As mentioned before, this looks particularly nice in string diagram notation. A 1-morphism $B\xleftarrow f A$ is drawn as a vertical line labeled by $f$ bisecting a square with the left and right regions labeled by $B,A$ respectively. In the case that the 1-morphism is $1_A$ then you can remove the line and have a square labeled by $A$. Composition of $1$-morphisms $g\circ f$ is denoted by drawing 2 parallel vertical lines labelled $g,f$ respectively, trisecting a square, labeling regions according to source and targets of $C\xleftarrow g B\xleftarrow f A$ respectively. $2$-morphisms are drawn by putting a dot on a line bisecting a square. The dot is labeled by the $2$-morphism, the lower and upper lines it connects by its source and target $1$-morphisms respectively and the left and right regions by their source and target objects. Horizontal composition is drawn by trisecting the square as before and vertical composition by stacking your squares on top of each other. In the case that the $2$-cell is the identity $1_f$, it is drawn exactly like the $1$-cell $f$.

In general, a $2$-cell $g_1\circ\cdots\circ g_n\xleftarrow \varphi f_1\circ\cdots\circ f_m$ is drawn by having $n,m$ lines connecting the bottom or top of a square respectively to a dot labeled by $\varphi$ and labeling lines and regions accordingly. In the case of our $2$-morphism $R\circ L\xleftarrow\eta 1_C$, we have three lines emanating from the dot, however $1_C$ was to be drawn without a line so we have only a two lines connecting at a dot in the middle of the square. We drop the dot and simply draw it as a cup with appropriate labels. Similarly, we can draw $\epsilon$ as a cap. Finally, one can see the identity above is simply a planar isotopy of curves, something intuitive from topology.

See the catsters for a better explanation: http://www.youtube.com/watch?v=pmvVE8AGAEA.


One of my friend was being confused with the same question as the one linked in your MSE post decided to ask Jean-Pierre Marquis the question (he is the author of the SEP article which you mentioned in a comment below the original post).

His reply was,

If you take arbitrary abstract categories and stipulate that a pair of adjoint functors exists between them, the only thing you can hold on to is their abstract or formal properties, e.g. the left adjoint preserving colimits, etc. Of course, in general, it is not the case that you have a forgetful functor, but the general point remains. (Recall that I used the case of the adjoint functor simply to illustrate the main general point.) One can and should consider a pair of adjoint functors as providing conceptual inverses. One has to be careful and look at the details, even more so since a functor can have both a left and a right adjoint! The best analogy is probably from topology with the notions of section and retraction to a given map. But again, one has to be careful and it is probably better to think about these up to homotopy.


Here is a physical example from what I consider to be one of the greatest books ever written, Sheaf Theory by Rosiak, on page-238.

Example 202 Both small-scale and large-scale projects, such as those in research or development, require resources. Resource allocation (through grants, investment funding, contracts, etc.) requires a detailed plan (for how those resources are to be spent), especially as the project increases in scale. If Rsrc is a category consisting of relevant resources, so that objects are resources (e.g., for simplicity, different-sized checks or bags of money) and morphisms are given by a natural relation between those resources (e.g., ≤ in the case of a uniform money-valuation of the different resource objects), and if ProjPlan is a category consisting of project tasks, given some natural ordering (e.g., by order of priority in the carrying out of the plan), then we might consider the functor

$$ V: \text{Rsrc} \to \text{Projplan} $$

that maps resource $r$ to the collection of project plans that are viable given that resource, and the functor

$$N : \text{Projplan} \to \text{Rsrc}$$

taking a project task p to all those resources that are necessary to complete the task (which, depending on how Rsrc is structured, say in a simple case of “costs,” might just amount to returning an interval bounded by the least cost for which the task could be carried out, and including all other more ample amounts)

We would probably not expect V and N to construct strict inverses to one another, for we do not expect that, for any given resource r, a list of necessary resources for those plans that are deemed viable given r would be equal to r. Though we might expect that, among the resources, r ≤ NV(r). Similarly, we would not expect that, for a given project task p, the result of applying N to p and then V to N(p), would always equal the same task p. Yet we would expect VN(p) ≤ p in ProjPlan. This suggests that we have an adjunction,

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  • $\begingroup$ Hopefully someone can clean up the latex and write the last picture in mathjax $\endgroup$ May 26 at 23:11

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