Defn: Let $\mathsf{C}$ be a locally small category which admits all small colimits. We say that $X \in \mathsf{C}$ is completely compact if the functor $\mathrm{Hom}_{\mathsf{C}}(X, \cdot): \mathsf{C} \rightarrow \mathsf{Set}$ preserves all small colimits.
Note that this is DIFFERENT from calling an object "compact", when we only required that $\mathrm{Hom}_{\mathsf{C}}(X, \cdot)$ preserves filtered colimits.
Let $\mathsf{C}$ be a category. A retract of an object $X \in \mathsf{C}$ is a tuple $(i,r,Y)$, where the composition
$$
Y \xrightarrow{i} X \xrightarrow{r} Y
$$
is $\mathrm{id}_Y$. We also say that $Y$ is a retract of $X$. We have the following fact for retracts:
FACT: $Y$ is both the equalizer and the coequalizer of the diagram
$$
X {{{i \circ r} \atop \longrightarrow}\atop{\longrightarrow \atop {\mathrm{id}_X}}} X.
$$
My question: How to show that if $X \in \mathsf{C}$ is completely compact, then any retract of $X$ is completely compact?
My attempts: Some discussions have been in the post
Why is a direct summand of a compact object compact? . Yet people only dealt with the compact case, but not completely compact case. In the compact case, we can prove the similar statement (replacing completely compact by compact everywhere) as follows:
Let $\alpha: \mathsf{I} \rightarrow \mathsf{C}$ be a functor (i.e. a diagram in $\mathsf{C}$), where $\mathsf{I}$ is filtered, then with the listed fact above, we have
\begin{align}
\varinjlim_{i \in \mathsf{I}} \mathrm{Hom}_{\mathsf{C}}(Y, \alpha(i)) &= \varinjlim_{i \in \mathsf{I}} \mathrm{Hom}_{\mathsf{C}}\left(\varinjlim_{j \in \mathsf{J} := \{0,1\} } X, \alpha(i)\right) \\
&= \varinjlim_{i \in \mathsf{I}} \varprojlim_{j \in \mathsf{J} := \{0,1\} } \mathrm{Hom}_{\mathsf{C}}( X, \alpha(i)) \\
&\cong \varprojlim_{j \in \mathsf{J} := \{0,1\} } \varinjlim_{i \in \mathsf{I}} \mathrm{Hom}_{\mathsf{C}}( X, \alpha(i)) \\
&\cong \varprojlim_{j \in \mathsf{J} := \{0,1\} } \mathrm{Hom}_{\mathsf{C}}( X, \varinjlim_{i \in \mathsf{I}} \alpha(i)) \\
&\cong \mathrm{Hom}_{\mathsf{C}}\left(\varinjlim_{j \in \mathsf{J} := \{0,1\} } X, \varinjlim_{i \in \mathsf{I}} \alpha(i)\right) \\
&\cong \mathrm{Hom}_{\mathsf{C}}(Y, \varinjlim_{i \in \mathsf{I}} \alpha(i)).
\end{align}
On the third line, we interchanged the order of limits and colimits because filtered colimits commute with finite limits in the category of sets. (See post About a specific step in a proof of the fact that filtered colimits and finite limits commute in $\mathbf{Set}$ .) The fourth line holds since $X$ is compact.
However, the aobve process FAILED since in the completely compact case, we shall check for all small colimits, not just filtered ones. So the commutativity of "colimits" and "limits" failed. So how to generalize the proof to completely compact case, or are there any new proofs?
Any references are also welcome.
Thank you all for answering and commenting! 🙂
Best Answer
Hint: Retracts are stable under arbitrary functors. Deduce that $\mathrm{Hom}(Y,-)$ is a retract of a cocontinuous functor. Colimits of cocontinuous functors are cocontinuous.