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A right adjoint functor preserves limits. Dually a left adjoint functor preserves colimits. I often forget which is which. Of course, you can look up a book on category theory or use internet. But it's nice if there is a good mnemonic method to remember these facts.

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    $\begingroup$ I've seen people refer to this fact as RAPL (Right Adjoints Preserve Limits), which is silly enough to be memorable, at least to me. $\endgroup$ – Miha Habič Jan 5 '13 at 16:09
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    $\begingroup$ I simply remember the fact that " L eft adjoints preserve l imits" is wrong. $\endgroup$ – roman Oct 15 '13 at 7:38
  • $\begingroup$ I gave you a single sentence. Hope it helps. :) $\endgroup$ – user304051 Dec 8 '16 at 1:52
  • $\begingroup$ LAPC, similar to suggestion by @MihaHabič . Just looked up that it stands for Los Angeles Poker Classic, Police Commission, etc although I wasn't aware of either $\endgroup$ – usr0192 Mar 7 '20 at 19:34
  • $\begingroup$ @roman I just remember that " 'Right adjoints preserve limits' is rect", and " 'Left adjoints preserve colimits' is correct" $\endgroup$ – PrimeRibeyeDeal Sep 12 '20 at 17:41
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I remember this (and related facts) as follows: A left adjoint $F$ is characterized by morphisms on $F(x)$, and a colimit is characterized by morphisms on it. Dually, a right adjoint $G$ is characterized by morphisms into $G$, and a limit is characterized by morphisms into it. So basically I just reprove it all the time, after all it is only one line:

$(\mathrm{colim}_i F(x_i),-) = \mathrm{lim}_i (F(x_i),-) = \mathrm{lim}_i (x_i,G(-))=(\mathrm{colim}_i x_i,G(-))=(F(\mathrm{colim}_i x_i),-)$

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  • $\begingroup$ This is a very effective way to memorize it. I learned adjunctions by myself and have a puzzling point about this whole left/ right (co)limit commutativity. If I am not mistaken, right adjunctions are given by a left Kan extension which is by a colimit formula. It implies right adjoints must commute with colimits. What is the wrong with this argument? $\endgroup$ – Bumblebee Jun 24 '20 at 18:53
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The right thing to be preserved is the limit.

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    $\begingroup$ This is nice. I'm going to use this. Well done. $\endgroup$ – The Count Dec 8 '16 at 1:39
  • $\begingroup$ You should add: being a right adjoint implies preserving any limit (to prevent from confusing with having a right adjoint). $\endgroup$ – Watson Apr 27 '18 at 20:04
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The easiest way to remember which is which is to work through the proof that, say, left adjoints preserve colimits. Here's a quick sketch, with $F \dashv U$:

\begin{align} \textrm{Hom}(F \varinjlim A_\bullet, B) \cong \textrm{Hom}(\varinjlim A_\bullet, U B) & \cong \varprojlim \textrm{Hom}(A_\bullet, U B) \\ & \cong \varprojlim \textrm{Hom}(F A_\bullet, B) \cong \textrm{Hom}(\varinjlim F A_\bullet, B) \end{align}

Unfortunately there is no really good mnemonic in general because the use of left/right is inconsistent. For example:

  • Right adjoints preserve limits, so they are left exact.
  • Right derived functors are left Kan extensions (when working with derived categories).
  • Monomorphisms constitute the right class of an orthogonal factorisation system in regular categories, but they are preserved by left exact functors (and so by right adjoints).

In the end the only way to be sure about which is which is to remember whether the thing in question appears on the left or on the right in the diagram invoked in the definition. So, for example:

  • Left adjoints are called ‘left’ because they appear on the left of the $\to$ in the bijective correspondence $$\frac{F A \to B}{A \to U B}$$
  • Left exact functors are ‘left’ because they preserve left exact sequences, which are ‘left’ because they are the left part of a short exact sequence: $$0 \longrightarrow A' \longrightarrow A \longrightarrow A''$$
  • Left derived functors are ‘left’ because they extend an exact sequence to the left: $$\cdots \longrightarrow L^1 F A'' \longrightarrow F A' \longrightarrow F A \longrightarrow F A'' \longrightarrow 0$$
  • Left Kan extensions are ‘left’ because the functor that takes a functor to its left Kan extension is the left adjoint of the precomposition functor.
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  • $\begingroup$ In your proof above, why can you replace the colimit inside Hom with a limit outside it? $\endgroup$ – gen Jan 9 '19 at 13:12
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Just remember one particular instance of left and right adjoints, for example left adjoint $F$ to the forgetful functor $U$ from groups to sets. $F(X)$ is the free group on the set $X$.

The forgetful functor $U$ obviously preserves products but not coproducts, whereas $F$ obviously preserves coproducts but not products.

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    $\begingroup$ More simple, $- \times X : \mathsf{Set} \to \mathsf{Set}$ is left adjoint to $\mathrm{Hom}(X,-)$ and preserves colimits (for example $(A \cup B) \times X = A \times X \cup B \times X$), but no limits (for example $\star \times X \neq \star$). $\endgroup$ – Martin Brandenburg Jan 5 '13 at 18:25
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    $\begingroup$ I use the additional mnemonic "LEFT" and "FREE" are both 4 letters word ("RIGHT" and "FORGETFUL" aren't) $\endgroup$ – Romuald Jan 7 '13 at 10:51
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Here is how I do it.

Sometimes limits are called inverse limits and denoted by $\varprojlim$, whereas colimits are called direct limits and denoted by $\varinjlim$.

Now, a functor which has an adjoint in some direction, preserve the limits in the same direction.

Unravelling, a functor which has a left adjoint (i.e. which is a right adjoint) preserves limits pointing to the left, that is to say, it preserves inverse limits, or just limits if you prefer. Dually, a functor which has a right adjoint preserves limits pointing to the right which are direct limits, aka colimits.

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This is only difficult if you get confused by the following facts that are stupidly written in all the textbooks

  1. A limit of a particular diagram is a right Kan extension
  2. A functor assigning limits to diagrams is a right adjoint but a left Kan extension.

What happens if you apply a right adjoint to a limit is that you still have a limit, but if you apply the right adjoint to the limit-assigning functor you obtain a relative right adjoint, which is not a left Kan extension. The correct fact is this

  1. A functor assigning limits to diagrams is a relative right adjoint, i.e. an absolute right Kan lift, which is in general not a Kan extension!

How do you tell if a lift or extension is a right lift or extension? The natural transformation points away from the composite and toward the functor being lifted or extended. Then relative right adjoints, which are absolute right Kan lifts, preserve all right Kan extensions, which limits are an example of.

Because I've never seen this anywhere in the literature, below is the actual fact on which the preservation result relies on. Note that this works in any $2$-category.

$$\require{AMScd}\begin{CD} \mathcal A @>Y>> \mathcal E @>G>> \mathcal C\\ @VXVV \overset\epsilon\Rightarrow @| \overset{\epsilon'}\Leftarrow @VFVV\\ \mathcal B @>Z>> \mathcal E @>J>>\mathcal D \end{CD}$$ (Since amscd is inflexbile, for clarity the natural transformations are supposed to be $ZX\overset\epsilon\Rightarrow Y$ and $FG\overset{\epsilon'}\Rightarrow J$).

Lemma. Suppose that

  1. $JZX\overset{\epsilon_1}\Rightarrow JY$ is a right Kan extension of $\mathcal A\xrightarrow{JY}\mathcal D$ along $\mathcal A\xrightarrow{X}\mathcal B$. In our case $JZX\overset{\epsilon_1}\Rightarrow JY$ is $JZX\overset{J\epsilon}\Rightarrow JY$.
  2. $FGZ\overset{\epsilon_2}\Rightarrow JZ$ a right Kan lift of $\mathcal B\xrightarrow{JZ}\mathcal D$ along $\mathcal C\xrightarrow{F}\mathcal D$. In our case $FGZ\overset{\epsilon_2}\Rightarrow JZ$ is $FGZ\overset{\epsilon'Z}\Rightarrow JZ$.
  3. $FGY\overset{\epsilon_3}\Rightarrow JY$ is a right Kan lift of $\mathcal A\xrightarrow{JY}\mathcal D$ along $\mathcal C\xrightarrow{F}\mathcal D$. In our case $FGY\overset{\epsilon_3}\Rightarrow JY$ is $FGY\overset{\epsilon'Y}\Rightarrow JY$.

Then

  • The unique right extension $GZX\overset{\epsilon_4}\Rightarrow GY$ of $\mathcal A\xrightarrow{GY}\mathcal C$ along $\mathcal A\xrightarrow{X}\mathcal B$ such that the right lift $FGZX\overset{\epsilon_2X}\Rightarrow JZX\overset{\epsilon_1}\Rightarrow JY$ of $\mathcal A\xrightarrow{GZX}\mathcal C$ of $\mathcal A\xrightarrow{JY}\mathcal D$ factors as $FGZX\overset{F\epsilon_4}\Rightarrow FGY\overset{\epsilon_3}\Rightarrow JY$. In our case $GZX\overset{\epsilon_4}\Rightarrow GY$ is $GZX\overset{G\epsilon}\Rightarrow GY$.

Corollary 1. A right Kan lift $\mathcal E\xrightarrow{G}\mathcal C$ of $\mathcal E\xrightarrow{J}\mathcal D$ preserves a Kan extension $\mathcal B\xrightarrow{Z}\mathcal E$ of $\mathcal A\xrightarrow{Y}\mathcal E$ if both $\mathcal B\xrightarrow{Z}\mathcal C$ and $\mathcal A\xrightarrow{Y}\mathcal E$ respect the right Kan lift.

Corollary 2.. By definition the relative $\mathcal E\xrightarrow{G}\mathcal C$ is a relative right adjoint $F\dashv_JG$ if it is a right Kan lift of $\mathcal E\xrightarrow{J}\mathcal D$ along $\mathcal C\xrightarrow{F}\mathcal D$ that is absolute, i.e. respected by every functor to $\mathcal E$. Hence, a relative right adjoint $F\dashv_J G$ preserves all right Kan extensions that $J$ does.

Corolalry 3. Since a right adjoint is the adjoint relative to the identity functor, it preserves all right Kan extensions.

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With the right limit, you're left with the colimit.

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Both right adjoints and limits talk about how maps into an object behave, hence they commute.

I like this since not only it is a good mnemonic, but it actually contains the idea used in the proof.

Alternatively, limits naturally appear to the right of an arrow.

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