Right Adjoints Preserve Limits (RAPL)

A set-valued functor with a left adjoint is representable, and representables preserve limits, so any set-valued functor with a left adjoint preserves limits. The restriction to set-valued functors is unnecessary. The general statement, abbreviated RAPL, is used constantly, both positively (a right adjoint automatically preserves limits) and negatively (a functor that fails to preserve some limit cannot have a left adjoint).¹

The theorem

The dual statement has its own name: left adjoints preserve colimits (LAPC). A left adjoint is characterized by maps out of its values, and colimits are defined by maps out; the symmetry is exact.

The comparison map

Preserves limits has a precise canonical form. Suppose both $\lim D$ in $\mathcal{B}$ and $\lim (G\circ D)$ in $\mathcal{A}$ exist. Applying $G$ to the limit cone of $D$ gives a cone on $G\circ D$ with vertex $G(\lim D)$, which factors uniquely through the limit cone of $G\circ D$:

where $p_I$ and $q_I$ are the two families of projections. The functor $G$ preserves the limit of $D$ exactly when this comparison map $\kappa$ is an isomorphism. RAPL asserts that for a right adjoint, $\kappa$ is invertible for every diagram whose limit exists; the four-line proof above produces the inverse by transposing cones across the adjunction. The comparison map is also what a failed preservation looks like in practice: for the forgetful $U:\mathbf{Grp}\to \mathbf{Set}$ and a coproduct diagram, the corresponding co-comparison map $U(\mathbb{Z})+U(\mathbb{Z})\to U(\mathbb{Z}\ast \mathbb{Z})$ identifies the two identity elements and hits only the one-syllable words of the free product, missing every genuinely mixed word, nowhere near a bijection.

Consequences

Forgetful functors and algebra

Forgetful functors from categories of algebras ($\mathbf{Grp}$, ${\mathbf{Vect}}_k$, $\mathbf{Mon}$, $\mathbf{Ring}$) to $\mathbf{Set}$ have left adjoints — the free constructions — so RAPL says they preserve all limits. This matches the computations of the limits-in-concrete-categories lesson: the underlying set of a product of groups is the product of the underlying sets, the underlying set of an equalizer is the equalizer of the underlying functions. RAPL explains why the underlying-set method works.

The same functors rarely have right adjoints, and correspondingly they rarely preserve colimits. The underlying set of a coproduct of groups (the free product) is far larger than the disjoint union of the underlying sets.²

Arithmetic in Set

Every set $B$ gives an adjunction $(-\times B)⊣(-{)}^B$ on $\mathbf{Set}$: functions $A\times B\to C$ correspond to functions $A\to C^B$. RAPL and LAPC applied to this pair produce the exponent laws of arithmetic.³

• $-\times B$ is a left adjoint, so it preserves colimits, in particular finite sums: $$\varnothing \times B\cong \varnothing ,(A_1+A_2)\times B\cong (A_1\times B)+(A_2\times B).$$
• $(-{)}^B$ is a right adjoint, so it preserves limits, in particular finite products: $$1^B\cong 1,(A_1\times A_2{)}^B\cong A_1^B\times A_2^B.$$

Taking cardinalities of finite sets recovers $0\cdot b=0$, distributivity, $1^b=1$, and $(a_1{a}_2{)}^b=a_1^b{a}_2^b$. The natural numbers are the isomorphism classes of finite sets, so these arithmetic identities are instances of RAPL and LAPC.

Commutation of limits via RAPL

When $\mathcal{A}$ has all limits of shape $\mathcal{I}$, the adjunction $\Delta ⊣\lim$ makes $\lim :[\mathcal{I},\mathcal{A}]\to \mathcal{A}$ a right adjoint. RAPL applied to it says $\lim$ preserves limits, which says that limits of shape $\mathcal{I}$ commute with all limits. This re-proves the Fubini-style commutation result for categories with all limits of one shape, with no computation.⁴

RAPL as a non-existence test

The contrapositive is a standard test: if a functor fails to preserve even one colimit, it has no right adjoint; if it fails to preserve one limit, it has no left adjoint.

Suppose $F:\mathbf{Set}\to \mathbf{Field}$ were left adjoint to $U$. Left adjoints preserve colimits, and the empty set is the initial object of $\mathbf{Set}$ (the empty colimit), so $F(\varnothing )$ would be an initial field: a field with exactly one homomorphism to every field. No such field exists. Field homomorphisms are injective and preserve $0\ne 1$, so any map of fields preserves characteristic; a single field cannot map to both $\mathbb{Q}$ (characteristic $0$) and ${\mathbb{F}}_p$ (characteristic $p$). Hence $\mathbf{Field}$ has no initial object and $F$ cannot exist. There is no free field on a set.⁵

The same template settles many cases.

• $U:\mathbf{Grp}\to \mathbf{Set}$ has no right adjoint. A right adjoint would make $U$ preserve colimits, but $U$ sends the coproduct $\mathbb{Z}\ast \mathbb{Z}$ (free product) to a set much bigger than $U(\mathbb{Z})+U(\mathbb{Z})$.
• The Yoneda embedding has no left adjoint in general. It fails to preserve colimits — the initial object goes to a non-initial presheaf — so it cannot be a right adjoint. (Preserving limits is consistent with being a right adjoint but does not prove it; the failure direction is the decisive one.)
• A functor between posets with a lower adjoint preserves meets. For order-preserving maps viewed as functors between thin categories, limits are meets and RAPL reads: a map with a lower (left) adjoint preserves all meets that exist. This is the classical Galois-connection fact.

Worked example (Floor and ceiling around $\mathbb{Z}\hookrightarrow \mathbb{R}$). Let $\iota :\mathbb{Z}\hookrightarrow \mathbb{R}$ be the inclusion, both sets ordered as usual. The ceiling is left adjoint to $\iota$ and the floor is right adjoint:

$$\lceil x\rceil \le n\leftrightarrow x\le \iota (n),n\le \lfloor x\rfloor \leftrightarrow \iota (n)\le x.$$

By RAPL and LAPC the floor (a right adjoint) preserves every meet that exists and the ceiling (a left adjoint) preserves every join. Finite meets and joins are uninformative in a total order, where $\min (x,y)$ is one of $x$ and $y$, so any monotone map preserves them. The content appears for infinite families. The floor preserves the infimum of any bounded family,

$$\lfloor \underset{n}{\inf }{x}_n\rfloor =\underset{n}{\inf }\lfloor x_n\rfloor ,$$

as RAPL requires. The ceiling does not: with $x_n=1+\frac{1}{n}$,

$$\lceil \underset{n}{\inf }(1+\frac{1}{n})\rceil =\lceil 1\rceil =1,\underset{n}{\inf }\lceil 1+\frac{1}{n}\rceil =\underset{n}{\inf }2=2,$$

so the ceiling fails to preserve an infinite meet, consistent with its being only a left adjoint. Dually the ceiling preserves suprema and the floor fails on them.

The inconclusive branch is real: preserving limits does not by itself produce a left adjoint. The unique functor $\mathcal{B}\to 1$ preserves every limit, yet it has a left adjoint exactly when $\mathcal{B}$ has an initial object, which plenty of complete-looking categories lack. The extra hypotheses under which preservation does suffice are the subject of the adjoint functor theorems.

Scope of the theorem

• RAPL preserves; it does not create. If $\mathcal{B}$ lacks the limit of $D$, RAPL says nothing about a limit of $G\circ D$ appearing in $\mathcal{A}$. Compare creation of limits, which is a stronger property enjoyed by, for example, forgetful functors of algebras.
• The converse fails without hypotheses. Limit preservation is necessary for being a right adjoint, not sufficient (the $\mathcal{B}\to 1$ example).
• Each adjoint preserves only its own side. A right adjoint may destroy colimits and a left adjoint may destroy limits; free functors typically destroy products (the free monoid on a one-element set is infinite, while the free monoid functor applied to a terminal object would need to be terminal to preserve it).

The converse direction — when limit preservation on $\mathcal{B}$ and $G$ guarantees a left adjoint — is settled by the general adjoint functor theorem, a size condition whose proof runs through the comma-category description of adjunctions via universal arrows.