Why is a direct summand of a compact object compact?

Question

In an additive category, we say that an object $A$ is compact if the functor $\text{Hom}(A, -)$ respects coproducts. That is, if the canonical morphism $$ \coprod_{i} \text{Hom} \left( A, X_{i} \right) \longrightarrow \text{Hom} \left( A, \coprod_{i} X_{i} \right) $$ is a bijection. Suppose $A \oplus B$ is compact. Why are the summands $A$ and $B$ compact? Everywhere claims this is obvious and provides no justification, but I cannot see why this is true.

Answer 1

More generally, a retract of a compact object is compact. Recall that expressing $U$ as a retraction of $V$ is to give morphisms

$$ U \xrightarrow{i} V \xrightarrow{p} U $$

whose composite is the identity on $U$. This is also called a split idempotent, since $ip$ is an idempotent map $V \to V$.

Note that any functor applied to a retract diagram gives another retract diagram.

The slick way to carry out the proof is to use the fact that the retract $U$ is both the equalizer and the coequalizer of the parallel maps $1_V$ and $ip$ (by the maps $i$ and $p$ respectively), and apply the general facts

$$ \hom(\mathop{\mathrm{colim}}_{i \in I} X_i , Y) \cong \lim_{i \in I} \hom(X_i, Y) $$ $$ \mathop{\mathrm{colim}}_{i \in I} \mathop{\mathrm{colim}}_{j \in J} X_{i,j} \cong \mathop{\mathrm{colim}}_{j \in J} \mathop{\mathrm{colim}}_{i \in I} X_{i,j} $$

Answer 2

You have a split exact sequence $0\to A\to A\oplus B\to B\to 0$ which produces the commutative diagram with split exact rows $$\require{AMScd} \begin{CD} 0 @>>> \coprod_{i} \text{Hom} \left( A, X_{i} \right) @>>> \coprod_{i} \text{Hom} \left( A\oplus B, X_{i} \right) @>>> \coprod_{i} \text{Hom} \left( B, X_{i} \right) @>>> 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> \text{Hom} \left( A, \coprod_{i} X_{i} \right) @>>> \text{Hom} \left( A\oplus B, \coprod_{i} X_{i} \right) @>>> \text{Hom} \left( B, \coprod_{i} X_{i} \right) @>>> 0 \end{CD}$$ where the middle vertical row is an isomorphism by assumption. Consequently, the left vertical arrow is monic, while the right vertical arrow is epic. By exchanging the roles of $A$ and $B$, we prove that these arrows are, in fact, isomorphisms.

Answer 3

In what follows, I shall add some detail to user14972's answer from the perspective of retract. Some more details and conventions are introduced in my question (post) Complete compactness and retract .

Let $\alpha: \mathsf{I} \rightarrow \mathsf{C}$ be a functor (i.e. a diagram in $\mathsf{C}$), where $\mathsf{I}$ is filtered, viewing the retract as a coequalizer (which is a special finite colimit), 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.