# nLab equalizer

### Context

#### Limits and colimits

limits and colimits

# Contents

## Definition

### In category theory

An equalizer is a limit

$eq\underset{\phantom{\rule{1em}{0ex}}e\phantom{\rule{1em}{0ex}}}{\to }x\underset{\phantom{\rule{1em}{0ex}}g\phantom{\rule{1em}{0ex}}}{\overset{f}{⇉}}y$\operatorname{eq}\underset{\quad e \quad}{\to}x\underoverset{\quad g \quad}{f}{\rightrightarrows}y

over a parallel pair i.e. of the diagram of the shape

$\left\{x\underset{\phantom{\rule{1em}{0ex}}g\phantom{\rule{1em}{0ex}}}{\overset{f}{⇉}}y\right\}\phantom{\rule{thinmathspace}{0ex}}.$\left\lbrace x \underoverset{\quad g \quad}{f}{\rightrightarrows} y \right\rbrace \,.

This means that for $f:x\to y$ and $g:x\to y$ two parallel morphisms in a category $C$, their equalizer is, if it exists

• an object $\mathrm{eq}\left(f,g\right)\in C$;

• a morphism $\mathrm{eq}\left(f,g\right)\to x$

• such that

• pulled back to $\mathrm{eq}\left(f,g\right)$ both morphisms become equal: $\left(\mathrm{eq}\left(f,g\right)\to x\stackrel{f}{\to }y\right)=\left(\mathrm{eq}\left(f,g\right)\to x\stackrel{g}{\to }y\right)$
• and $\mathrm{eq}\left(f,g\right)$ is the universal object with this property.

The dual concept is that of coequalizer.

### In type theory

In type theory the equalizer

$P\to A\stackrel{\stackrel{f}{\to }}{\underset{g}{\to }}B$P \to A \stackrel{\overset{f}{\to}}{\underset{g}{\to}} B

is given by the dependent sum over the dependent equality type

$P\simeq \sum _{a:A}\left(f\left(a\right)=g\left(a\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$P \simeq \sum_{a : A} (f(a) = g(a)) \,.

## Examples

• In $C=$ Set the equalizer of two functions of sets is the subset of elements of $c$ on which both functions coincide.

$\mathrm{eq}\left(f,g\right)=\left\{s\in c\mid f\left(s\right)=g\left(s\right)\right\}\phantom{\rule{thinmathspace}{0ex}}.$eq(f,g) = \left\{ s \in c | f(s) = g(s) \right\} \,.
• For $C$ a category with zero object the equalizer of a morphism $f:c\to d$ with the corresponding zero morphism is the kernel of $f$.

## Properties

###### Proposition

A category has equalizers if it has products and pullbacks.

###### Proof

For $S\stackrel{\stackrel{g}{\to }}{\underset{f}{\to }}T$ the given diagram, first form the pullback

$\begin{array}{ccc}S{×}_{f,g}S& \to & S\\ ↓& & {↓}^{g}\\ S& \stackrel{f}{\to }& T\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ S \times_{f,g} S &\to& S \\ \downarrow && \downarrow^{\mathrlap{g}} \\ S &\stackrel{f}{\to}& T } \,.

This gives a morphism $S{×}_{f,g}S\to S×S$ into the product.

Define $\mathrm{eq}\left(f,g\right)$ to be the further pullback

$\begin{array}{ccc}\mathrm{eq}\left(f,g\right)& \to & S{×}_{f,g}S\\ ↓& & ↓\\ S& \stackrel{\left(\mathrm{id},\mathrm{id}\right)}{\to }& S×S\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ eq(f,g) &\to& S \times_{f,g} S \\ \downarrow && \downarrow \\ S &\stackrel{(id, id)}{\to}& S \times S } \,.

One checks that the vertical morphism $\mathrm{eq}\left(f,g\right)\to S$ equalizes $f$ and $g$ and that it does so universally.

###### Proposition

If a category has products and equalizers, then it has limits; see there.

Revised on November 15, 2012 13:55:18 by Stephan Alexander Spahn (79.227.160.54)