Topos(2): Equalizer

MotivationEqualizerEqualizer and SubobjectExample of Equalizers in Different Categories

Let Focalors and Furina be two maps from the position of the Hydro Archon to the image in the people's hearts.

What is the equalizer of Focalors and Furina?

Motivation

We always encounter the symbol $=$$=$ in our lives. When we were kids, we were asked to solve equations like $f(x)=g(x)$$f(x)=g(x)$ in $\mathrm{Set}$$\mathrm{Set}$.

What is the answer? The equalizer of $f$$f$ and $g$$g$! This idea can be generalized to some category $\mathcal{C}\ne \mathrm{Set}$$\mathcal{C} \ne \mathrm{Set}$ as well.

Equalizer

Definition.

Given a category $\mathcal{C}$$\mathcal{C}$ and parallel arrows $f,g:X\to Y$$f, g: X \to Y$, then the derived category of forks ${\mathcal{C}}_{f||g}$$\mathcal{C}_{f||g}$ has objects as forks $(W,k)$$(W,k)$.

The morphism between two objects $(W^{\prime },k^{\prime })$$(W',k')$ and $(W,k)$$(W,k)$ is a $\mathcal{C}$$\mathcal{C}$-arrow $v$$v$ such that $k^{\prime }=k\circ v$$k' = k \circ v$.

The final object $(E,e)$$(E,e)$ in ${\mathcal{C}}_{f||g}$$\mathcal{C}_{f||g}$, if it exists, will be called the equalizer of $f$$f$ and $g$$g$. By definition, the equalizer is unique up to isomorphism.

The definition tells us that for any $(W,k)$$(W,k)$ that equalizes $f$$f$ and $g$$g$, i.e. $f\circ k=g\circ k$$f \circ k = g \circ k$, it has to factor through $(E,e)$$(E,e)$ via the unique $u$$u$.

Many problems we encounter in mathematics ask: what is the equalizer of these two morphisms?

Consider a high school problem as follows.

Let $\mathcal{C}=\mathrm{Set}$$\mathcal{C} = \mathrm{Set}$, $f=\sin x$$f = \sin x$, $g=\cos x$$g = \cos x$, $X=Y=\mathbb{R}$$X = Y = \mathbb{R}$. Then what is the equalizer of $f$$f$ and $g$$g$?

Notice that for any $(W,k)$$(W,k)$ such that $\sin (k(x))=\cos (k(x))$$\sin(k(x)) = \cos(k(x))$, the $k(x)$$k(x)$ has to lie in the set $\{\frac{\pi }{4}+k\pi \mid k\in \mathbb{Z}\}$$\{\frac{\pi}{4} + k\pi \mid k \in \mathbb{Z}\}$.

Hence, let $E=\{\frac{\pi }{4}+k\pi \mid k\in \mathbb{Z}\}$$E = \{\frac{\pi}{4} + k\pi \mid k \in \mathbb{Z}\}$ and $E\stackrel{e}{\hookrightarrow }\mathbb{R}$$E \xhookrightarrow{e} \mathbb{R}$ be the inclusion map. It is easy to see that $(E,e)$$(E,e)$ satisfies the universal property of the equalizer. Here, $(E,e)$$(E,e)$ is a subobject of $X$$X$ since $e$$e$ is a monomorphism. Is this true in general? The answer is yes! Let's prove it.

Equalizer and Subobject

Proposition. The equalizer of $f$$f$ and $g$$g$ is always a subobject of $X$$X$.

Proof. By the definition of a subobject, we only need to prove that $e:E\hookrightarrow X$$e: E \hookrightarrow X$ is a monomorphism.

Let $h,h^{\prime }$$h, h'$ be two morphisms such that $e\circ h=e\circ h^{\prime }$$e \circ h = e \circ h'$, denote it as $k$$k$. Then $f\circ k=g\circ k$$f \circ k = g \circ k$. By the universal property of $(E,e)$$(E,e)$, there exists a unique $u$$u$ such that $k=e\circ u$$k = e \circ u$. Hence $h=u=h^{\prime }$$h = u = h'$. Thus, $e\circ h=e\circ h^{\prime }⟹h=h^{\prime }$$e \circ h = e \circ h' \implies h = h'$, which tells us $e$$e$ is a monomorphism, and $(E,e)$$(E,e)$ is a subobject of $X$$X$. $◻$$\square$

Proposition. Let $(E,e)$$(E,e)$ be an equalizer of $f$$f$ and $g$$g$, and $(E^{\prime },e^{\prime })\cong (E,e)$$(E',e') \cong (E,e)$. Then $(E^{\prime },e^{\prime })$$(E',e')$ is an equalizer of $f$$f$ and $g$$g$ as well. Conversely, if $(E^{\prime },e^{\prime })$$(E',e')$ is another equalizer, then $(E^{\prime },e^{\prime })\cong (E,e)$$(E',e') \cong (E,e)$.

Proof. Let $(E^{\prime },e^{\prime })$$(E',e')$ be a subobject of $X$$X$ and $(E^{\prime },e^{\prime })\cong (E,e)$$(E',e') \cong (E,e)$. First, we need to prove that $f\circ e^{\prime }=g\circ e^{\prime }$$f \circ e' = g \circ e'$.

Notice that $e\circ h=e^{\prime }$$e \circ h = e'$, hence $f\circ e^{\prime }=(f\circ e)\circ h=(g\circ e)\circ h=g\circ e^{\prime }$$f \circ e' = (f \circ e) \circ h = (g \circ e) \circ h = g \circ e'$, hence $f\circ e^{\prime }=g\circ e^{\prime }$$f \circ e' = g \circ e'$. Thus, $(E,e)\cong (E^{\prime },e^{\prime })$$(E,e) \cong (E',e')$ in ${\mathcal{C}}_{f||g}$$\mathcal{C}_{f||g}$. Hence $(E^{\prime },e^{\prime })$$(E',e')$ is a final object as well. The converse is obvious. $◻$$\square$

Example of Equalizers in Different Categories

In $\mathrm{Set}$$\mathrm{Set}$, an equalizer of $f$$f$ and $g$$g$ is the subset $(\{x\in X\mid f(x)=g(x)\},i)$$(\{x \in X \mid f(x) = g(x)\}, i)$.

Well, the equalizer in concrete categories whose objects are sets with structure behaves similarly.

But the category of sets has another good property. Let $2=\mathbb{Z}/2\mathbb{Z}$$2 = \mathbb{Z}/2\mathbb{Z}$, and $S$$S$ be a subset of $X$$X$, $\chi$$\chi$ be the characteristic function of $S$$S$, and $1_X$$1_X$ be the constant function such that $1_X(x)=1$$1_X(x) = 1$ for all $x\in X$$x \in X$.

Then $S$$S$ is the equalizer of $1_X$$1_X$ and $\chi$$\chi$.

Let $\mathrm{Mon}$$\mathrm{Mon}$ be the category of monoids and consider the equalizer of two monoid homomorphisms $f,g:X\to Y$$f, g: X \to Y$. As a set, it should be

$$\begin{array}{}\text{(1)}& (E,e)=(\{x\in X\mid f(x)=g(x)\},i).\end{array}$$

But, is that a monoid?

Firstly, it is easy to see that the identity element $1\in E$$1 \in E$. Secondly, if $a,b\in E$$a, b \in E$, then $f(ab)=f(a)f(b)=g(a)g(b)=g(ab)$$f(ab) = f(a)f(b) = g(a)g(b) = g(ab)$. Hence $a,b\in E⟹ab\in E$$a, b \in E \implies ab \in E$ as well. Thus, $E$$E$ is a monoid.

Let $\mathrm{Mod}(R)$$\mathrm{Mod}(R)$ be the category of $R$$R$-modules. Then the equalizer of $f,g:X\to Y$$f, g: X \to Y$ is just $(\ker (f-g),i)$$(\mathrm{ker}(f-g), i)$.

Let's check that. Let $k$$k$ be an $R$$R$-module homomorphism such that $f\circ k=g\circ k$$f \circ k = g \circ k$. Then $f(k(x))=g(k(x))$$f(k(x)) = g(k(x))$, i.e., $(f-g)(k(x))=0$$(f-g)(k(x)) = 0$. Hence, the image of $k$$k$ is in the kernel of $f-g$$f-g$ and the unique $u$$u$ is just $k:W\to \mathrm{Im}(k)\subseteq \ker (f-g)$$k: W \to \mathrm{Im}(k) \subseteq \mathrm{ker}(f-g)$.

Remark. The kernel of an $R$$R$-module homomorphism $f$$f$ is just the equalizer of $f$$f$ and $0$$0$. You can do the same thing for the category of groups.

Let $\mathrm{Pos}$$\mathrm{Pos}$ be the category of posets, then the equalizer of $f,g:X\to Y$$f, g: X \to Y$ is $(\{x\in X\mid f(x)=g(x)\},i)$$(\{x \in X \mid f(x) = g(x)\}, i)$ with the inherent order.

Let $\mathrm{Top}$$\mathrm{Top}$ be the category of topological spaces, then the equalizer of $f,g:X\to Y$$f, g: X \to Y$ is $(\{x\in X\mid f(x)=g(x)\},i)$$(\{x \in X \mid f(x) = g(x)\}, i)$ with the subspace topology. It is easy to see the reason that $E$$E$ is equipped with the subspace topology. Let $(E,e)$$(E,e)$ be equipped with another topology making $e$$e$ continuous. Let $(W,k)$$(W,k)$ be the set $E$$E$ with the subspace topology and $e$$e$ be the inclusion map. Then there is no continuous $u$$u$ such that the following diagram commutes.