Mnemonic for the fact that a right(left) adjoint functor preserves limits(colimits)

Question

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.

Answer 1

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.

Answer 2

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:

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

Answer 3

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.