# Kerodon

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### 8.4.2 Idempotents in Ordinary Categories

Let $M$ be a monoid. Recall that an element $e \in M$ is idempotent if it satisfies the equation $e^2 = e$. We now consider the special case where $M = \operatorname{End}_{\operatorname{\mathcal{C}}}(X) = \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,X)$ is the set of endomorphisms of an object $X$ of some category $\operatorname{\mathcal{C}}$.

Definition 8.4.2.1. Let $\operatorname{\mathcal{C}}$ be a category. An idempotent endomorphism in $\operatorname{\mathcal{C}}$ is a pair $(X,e)$, where $X$ is an object of $\operatorname{\mathcal{C}}$ and $e: X \rightarrow X$ is an endomorphism of $X$ which satisfies the identity $e = e \circ e$, so that we have a commutative diagram

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{e} & \\ X \ar [ur]^{e} \ar [rr]^{e} & & X. }$

In this situation, we will also say that $e$ is an idempotent endomorphism of $X$.

Example 8.4.2.2 (Identity Morphisms). Let $\operatorname{\mathcal{C}}$ be a category. For every object $X \in \operatorname{\mathcal{C}}$, the identity morphism $\operatorname{id}_{X}: X \rightarrow X$ is an idempotent endomorphism in $\operatorname{\mathcal{C}}$. Conversely, if $e: X \rightarrow X$ is an idempotent endomorphism in $\operatorname{\mathcal{C}}$ which is also an isomorphism, then $e = \operatorname{id}_{X}$.

Example 8.4.2.3 (Split Idempotents). Let $\operatorname{\mathcal{C}}$ be a category containing a retraction diagram

8.27
$$\begin{gathered}\label{equation:retracts-and-idempotents} \xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{r} & \\ Y \ar [ur]^{i} \ar [rr]^{ \operatorname{id}_{Y} } & & Y } \end{gathered}$$

(see Definition 8.4.1.16). Then $e = i \circ r$ is an idempotent endomorphism of $X$. This follows from the calculation

$e \circ e = (i \circ r) \circ (i \circ r) = i \circ \operatorname{id}_{Y} \circ r = i \circ r = e.$

We will say that an idempotent endomorphism $e: X \rightarrow X$ is split if it can be obtained in this way (that is, if $e = i \circ r$, for some pair of morphisms $i: Y \rightarrow X$ an $r: X \rightarrow Y$ satisfying $r \circ i = \operatorname{id}_{Y}$.

In the situation of Example 8.4.2.3, the diagram (8.27) can be recovered (up to isomorphism) from the idempotent endomorphism $e: X \rightarrow X$, by virtue of the following:

Proposition 8.4.2.4. Let $\operatorname{\mathcal{C}}$ be a category containing a retraction diagram

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{r} & \\ Y \ar [ur]^{i} \ar [rr]^{ \operatorname{id}_{Y} } & & Y, }$

and let $e = i \circ r$ be the idempotent endomorphism Example 8.4.2.3. Then:

$(1)$

The morphism $i$ exhibits $Y$ as an equalizer of the pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$.

$(2)$

The morphism $r$ exhibits $Y$ as a coequalizer of the pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$.

Proof. We will prove $(1)$; the proof of $(2)$ is similar. Fix an object $Z \in \operatorname{\mathcal{C}}$ and a morphism $f: Z \rightarrow X$ satisfying $e \circ f = \operatorname{id}_{X} \circ f$; we wish to show that there is a unique morphism $g: Z \rightarrow Y$ satisfying $i \circ g = f$. To prove uniqueness, we note that $g$ is determined by the identity

$g = \operatorname{id}_{Y} \circ g = (r \circ i) \circ g = r \circ (i \circ g) = r \circ f.$

To establish existence, we observe that the composition $g = r \circ f$ satisfies the identity

$i \circ g = i \circ (r \circ f) = (i \circ r) \circ f = e \circ f = \operatorname{id}_{X} \circ f = f.$
$\square$

Corollary 8.4.2.5. Let $\operatorname{\mathcal{C}}$ be a category and let $e: X \rightarrow X$ be an idempotent endomorphism in $\operatorname{\mathcal{C}}$. The following conditions are equivalent:

$(1)$

The idempotent endomorphism $e$ splits. That is, $e$ admits a factorization $X \xrightarrow {r} Y \xrightarrow {i} X$, where $r \circ i = \operatorname{id}_{Y}$.

$(2)$

The pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$ admit an equalizer in $\operatorname{\mathcal{C}}$.

$(3)$

The pair of morphisms $(e, \operatorname{id}_{X}): X \rightrightarrows X$ admit a coequalizer in $\operatorname{\mathcal{C}}$.

Proof. The implications $(1) \Rightarrow (2)$ and $(1) \Rightarrow (3)$ follow from Proposition 8.4.2.4. We will show that $(2)$ implies $(1)$; the proof of the implication $(3) \Rightarrow (1)$ is similar. Suppose that there exists a morphism $i: Y \rightarrow X$ which exhibits $Y$ as an equalizer of the pair of morphisms $(e,\operatorname{id}_ X): X \rightrightarrows X$. Since $e$ is idempotent, we have $e \circ e = e = \operatorname{id}_{X} \circ e$. Invoking the universal property of $Y$, we deduce that there is a unique morphism $r: X \rightarrow Y$ satisfying $e = i \circ r$. To complete the proof, it will suffice to show that $r \circ i$ is the identity morphism from $Y$ to itself. Since $i$ is a monomorphism, this follows from the calculation

$i \circ (r \circ i) = (i \circ r) \circ i = e \circ i = \operatorname{id}_{X} \circ i = i = i \circ \operatorname{id}_{Y}.$
$\square$

Corollary 8.4.2.6. Let $\operatorname{\mathcal{C}}$ be a category which admits equalizers (or coequalizers). Then every idempotent endomorphism in $\operatorname{\mathcal{C}}$ is split.

Construction 8.4.2.7 (The Universal Idempotent). We define a category $\operatorname{Idem}$ as follows:

• The category $\operatorname{Idem}$ has single object $\widetilde{X}$.

• Morphisms in $\operatorname{Idem}$ are given by $\operatorname{Hom}_{\operatorname{Idem}}( \widetilde{X}, \widetilde{X}) = \{ \operatorname{id}_{\widetilde{X}}, \widetilde{e} \}$.

• The composition law on $\operatorname{Idem}$ is given (on non-identity morphisms) by $\widetilde{e} \circ \widetilde{e} = \widetilde{e}$.

Remark 8.4.2.8. Let $\operatorname{\mathcal{C}}$ be a category and let $e: X \rightarrow X$. be an idempotent endomorphism in $\operatorname{\mathcal{C}}$. Then there is a unique functor $F: \operatorname{Idem}\rightarrow \operatorname{\mathcal{C}}$ satisfying $F(\widetilde{X}) = X$ and $F( \widetilde{e}) = e$.

Remark 8.4.2.10. Let $\operatorname{Ret}$ denote the category introduced in Construction 8.4.0.1. Then $\operatorname{Idem}$ can be identified with the full subcategory of $\operatorname{Ret}$ spanned by the object $\widetilde{X}$. Let $\operatorname{\mathcal{C}}$ be a category and let $\overline{F}: \operatorname{Ret}\rightarrow \operatorname{\mathcal{C}}$ be the functor determined by a retraction diagram in $\operatorname{\mathcal{C}}$ (see Exercise 8.4.0.2). Then the restriction $F = \overline{F}|_{ \operatorname{Idem}}$ corresponds (under the identification of Remark 8.4.2.8) to the idempotent endomorphism of Example 8.4.2.3.