nLab
equalizer

Definition

An equalizer is a limit

$$eq\underset{e}{\to }x\underset{g}{\overset{f}{\rightrightarrows }}y$$\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

$$\left\{x\underset{g}{\overset{f}{\rightrightarrows }}y\right\}.$$\left\lbrace x \underoverset{\quad g \quad}{f}{\rightrightarrows} y \right\rbrace \,.

(See also fork diagram).

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 $\mathrm{eq}(f,g)\in C$;

• a morphism $\mathrm{eq}(f,g)\to x$

• such that

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

The dual concept is that of coequalizer.

Examples

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

  $$\mathrm{eq}(f,g)=\{s\in c\mid f(s)=g(s)\}.$$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

$$\begin{array}{ccc}S{\times }_{f,g}S& \to & S\\ \downarrow & & {\downarrow }^g\\ S& \stackrel{f}{\to }& T\end{array}.$$\array{ S \times_{f,g} S &\to& S \\ \downarrow && \downarrow^{\mathrlap{g}} \\ S &\stackrel{f}{\to}& T } \,.

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

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

$$\begin{array}{ccc}\mathrm{eq}(f,g)& \to & S{\times }_{f,g}S\\ \downarrow & & \downarrow \\ S& \stackrel{(\mathrm{id},\mathrm{id})}{\to }& S\times S\end{array}.$$\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 $\mathrm{eq}(f,g)\to S$ equalizes $f$ and $g$ and that it does so universally.