Topos (2): Subobject functor via pullback, subobject classifier.

PullbackExamples of pullback.Subobject functor and subobject classifier.Subobject functor.Subobject classifier, the first definition.Subobject classifier, the second definition.The equivalence of two definitions.

 

Pullback

Motivation.

In topos (1), we have defined what is subobject, in this section, we will define subobject functor via pullback square.

So what is pullback?

Consider

$$\begin{array}{}\text{(1)}& D_1\stackrel{d_1}{\to }{D}_3\stackrel{d_2}{\leftarrow }{D}_2\end{array}$$

There exists a $L$$L$ such that for any $c_1:C\to D_1,c_2:C\to D_2$$c_1:C\to D_1,c_2:C\to D_2$ such that $d_1\circ c_1=d_2\circ c_2$$d_1\circ c_1=d_2\circ c_2$, there exists a unique $u$$u$ make the following diagram commute:

Or, $L$$L$ is the soulution of the universal map.

This is a limit or you can think it as a final object in a proper category.

Some times we call $L$$L$ fiber product as well. Since if you consider two fibers over a topological space $B$$B$

$$\begin{array}{}\text{(2)}& {\pi }_X:X\to B,{\pi }_Y:Y\to B\end{array}$$

Their product will be the fiber product of:

$$\begin{array}{}\text{(3)}& X\stackrel{{\pi }_X}{\to }B\stackrel{{\pi }_Y}{\leftarrow }Y\end{array}$$

Examples of pullback.

In $\mathrm{Set}$$\mathrm{Set}$, Let $d_1\circ c_1(c)=d_2\circ c_2(c)$$d_1\circ c_1(c)=d_2\circ c_2(c)$ for all $c\in C$$c\in C$, then $d_1(c_1(c))=d_2(c_2(c))$$d_1(c_1(c))=d_2(c_2(c))$ for all $c\in C$$c\in C$.

Hence $⟨c_1(c),c_2(c)⟩\in D_1\times D_2$$\langle c_1(c),c_2(c)\rangle\in D_1\times D_2$ satisfies that $d_1(c_1(c))=d_2(c_2(c))$$d_1(c_1(c))=d_2(c_2(c))$. Hence for all $(C,c_1,c_2)$$(C,c_1,c_2)$, it will universally path through $D_1{\times }_{D_3}{D}_2:=\{⟨x,y⟩\in D_1\times D_2|d_1(x)=d_2(y)\}$$D_1\times_{D_{3}} D_2:=\{\langle x,y\rangle\in D_1\times D_2|d_1(x)=d_2(y)\}$ via $u(c)=⟨c_1(c),c_2(c)⟩$$u(c)=\langle c_1(c),c_2(c)\rangle$​.

Consider the following diagram, where $i_1,i_2$$i_1,i_2$ are inclusion map.

Then $L=\{⟨x,y⟩\in X\times Y|i_1(x)=i_2(y)\}\cong X\cap Y.$$L=\{\langle x,y\rangle\in X\times Y|i_1(x)=i_2(y)\}\cong X\cap Y.$

If we consider

Then $L=\{⟨x,y⟩|f(x)=y\}\cong f^{-1}(Y)$$L=\{\langle x,y\rangle|f(x)=y\}\cong f^{-1}(Y)$​

In general, if we consider the pullback of

$$\begin{array}{}\text{(4)}& X\stackrel{f}{\to }Y\stackrel{i}{\leftarrow }U\end{array}$$

Where $U\subseteq Y$$U\subseteq Y$ and $i$$i$ is the natural embedding. Then the result of pullback will be $f^{-1}(U)$$f^{-1}(U)$.

We will catch this idea later, and see how to use pullback to define subobject functor.

In $\mathrm{Ring}$$\mathrm{Ring}$, the pullback or fiber product exists.

Readers could check that

$$\begin{array}{}\text{(5)}& D_1{\times }_{D_3}{D}_2:=\{⟨x,y⟩\in D_1\times D_2|d_1(x)=d_2(y)\}\end{array}$$

is a subring of $D_1\times D_2$$D_1\times D_2$​.

Subobject functor and subobject classifier.

Proposition. Pulling back a monomorphism yields a monomorphism.

That is, let the following diagram be a pullback diagram, $(Y,g)$$(Y,g)$ is a subobject of $Z$$Z$ implies that $(L,a)$$(L,a)$ is a subobject of $X.$$X.$ i.e. Give you a morphism $f:X\to Z$$f:X\to Z$ and a subobject of $Z$$Z$, the pullback will gives you a subobject of $X$$X$. We will use this to define subobject functor.

Proof.

Consider $j,k:C\to L$$j,k:C\to L$ such that $a\circ j=a\circ k$$a\circ j=a\circ k$. We want to show that $j=k$$j=k$ via to show that $b\circ j=b\circ k$$b\circ j=b\circ k$.

If $b\circ j=b\circ k$$b\circ j=b\circ k$, then this two cone over $L$$L$ are same hence $j=k$$j=k$ since there exists a unique morphism such that the diagram commute by the definition of pullback.

We know that $g\circ b\circ j=f\circ a\circ j=f\circ a\circ k=g\circ b\circ k$$g\circ b\circ j=f\circ a\circ j=f\circ a\circ k=g\circ b\circ k$. Since $g$$g$ is mono,$b\circ j=b\circ k$$b\circ j=b\circ k$, hence $j=k$$j=k$, hence $a$$a$ is momo as well, hence $(L,a)$$(L,a)$ is a subobject of $X$$X$.$◻$$\square$

Subobject functor.

Let $\mathcal{C}$$\mathcal C$ be a category such that the pullback exists. Suppose that for any $A\in \mathrm{Ob}(\mathcal{C}),{\mathrm{Sub}}_{\mathcal{C}}(A)$$A\in\mathrm{Ob}(\mathcal C),\mathrm{Sub}_{\mathcal C}(A)$ is a set, we call $\mathcal{C}$$\mathcal C$ well powered category. Clearly $\mathrm{Set}$$\mathrm{Set}$ is well powered category.

Then we get a functor

$$\begin{array}{}\text{(6)}& {\mathrm{Sub}}_{\mathcal{C}}(-):{\mathcal{C}}^{\mathrm{op}}\to \mathrm{Set}\end{array}$$

For an object $Z$$Z$ in $\mathcal{C}$$\mathcal C$, it will be mapped to ${\mathrm{Sub}}_{\mathcal{C}}(Z)$$\mathrm{Sub}_{\mathcal C}(Z)$. For a morphism $f:X\to Z$$f:X\to Z$, the previous propositions shows us that the pullback will map each subobject $(Y,g)$$(Y,g)$ to a unique $(L,a)$$(L,a)$ (uo to isomorphism).

The map ${\mathrm{Sub}}_{\mathcal{C}}(f):{\mathrm{Sub}}_{\mathcal{C}}(Z)\to {\mathrm{Sub}}_{\mathcal{C}}(X)$$\mathrm{Sub}_{\mathcal C}(f):\mathrm{Sub}_{\mathcal C}(Z)\to\mathrm{Sub}_{\mathcal C}(X)$ is well defined. Let $(L^{\prime },a^{\prime })$$(L',a')$ be another pullback, then it will be isomorphic to $(L,a)$$(L,a)$ as a subobject.

Subobject classifier, the first definition.

If the functor ${\mathrm{Sub}}_{\mathcal{C}}(-)$$\mathrm{Sub}_{\mathcal C}(-)$ is representable, i.e. ${\mathrm{Sub}}_{\mathcal{C}}(-)\cong {\mathrm{Hom}}_{\mathcal{C}}(-,\mathrm{\Omega })$$\mathrm{Sub}_{\mathcal C}(-)\cong\mathrm{Hom}_{\mathcal C}(-,\Omega)$ then we say $\mathrm{\Omega }$$\Omega$​​​ is the subobject classifier.

Hence we have $X⟼{\mathrm{Hom}}_{\mathcal{C}}(X,\mathrm{\Omega }),(f:X\to Y)⟼f^{\ast }:{\mathrm{Hom}}_{\mathcal{C}}(Y,\mathrm{\Omega })\to {\mathrm{Hom}}_{\mathcal{C}}(X,\mathrm{\Omega })$$X\longmapsto \mathrm{Hom}_{\mathcal C}(X,\Omega),(f:X\to Y)\longmapsto f^*:\mathrm{Hom}_{\mathcal C}(Y,\Omega)\to\mathrm{Hom}_{\mathcal C}(X,\Omega)$.

This is interesting since we call $f^{\ast }$$f^*$ pullback as well.

The subobject classifier is unique up to isomorphism via Yoneda lemma.

Also, if $\mathcal{C}$$\mathcal C$ is locally small and ${\mathrm{Sub}}_{\mathcal{C}}(-)\cong {\mathrm{Hom}}_{\mathcal{C}}(-,\mathrm{\Omega })$$\mathrm{Sub}_{\mathcal C}(-)\cong\mathrm{Hom}_{\mathcal C}(-,\Omega)$, then $\mathcal{C}$$\mathcal C$ is well powered since ${\mathrm{Sub}}_{\mathcal{C}}(X)\cong {\mathrm{Hom}}_{\mathcal{C}}(X,\mathrm{\Omega })$$\mathrm{Sub}_{\mathcal C}(X)\cong\mathrm{Hom}_{\mathcal C}(X,\Omega)$​ is a set.

If we consider the Yoneda embedding $\mathcal{C}\to {\mathrm{Set}}^{{\mathcal{C}}^{\mathrm{op}}}$$\mathcal C\to\mathrm{Set}^{\mathcal C^{\mathrm{op}}}$, then the image of $\mathrm{\Omega }$$\Omega$ will be ${\mathrm{Hom}}_{\mathcal{C}}(-,\mathrm{\Omega })$$\mathrm{Hom}_{\mathcal C}(-,\Omega)$, hence ${\mathrm{Sub}}_{\mathcal{C}}(-)$$\mathrm{Sub}_{\mathcal C}(-)$​ is the subobject classifier of the image of the Yoneda embedding when the subobject classifier exists.

Example. As we know, when $\mathcal{C}=\mathrm{Set}$$\mathcal C=\mathrm{Set}$, the subobject functor is just the power set functor $P:{\mathrm{Set}}^{\mathrm{op}}\to \mathrm{Set}$$P:\mathrm{Set}^{\mathrm{op}}\to\mathrm{Set}$.

$$\begin{array}{}\text{(7)}& X\stackrel{f}{\to }Y\stackrel{i}{\leftarrow }U\end{array}$$

Where $U\subseteq Y$$U\subseteq Y$ and $i$$i$ is the natural embedding. Then the result of pullback will be $f^{-1}(U)$$f^{-1}(U)$.

The power set functor is represented by $\mathbb{Z}/2\mathbb{Z}$$\mathbb Z/2\mathbb Z$. i.e. ${\mathrm{Sub}}_{\mathrm{Set}}(-)\cong {\mathrm{Hom}}_{\mathrm{Set}}(-,\mathbb{Z}/2\mathbb{Z})$$\mathrm{Sub}_{\mathrm{Set}}(-)\cong\mathrm{Hom}_{\mathrm{Set}}(-,\mathbb Z/2\mathbb Z)$. Hence $\mathbb{Z}/2\mathbb{Z}$$\mathbb Z/2\mathbb Z$ is the object classifier in category of set. The elements in ${\mathrm{Hom}}_{\mathrm{Set}}(X,\mathbb{Z}/2\mathbb{Z})$$\mathrm{Hom}_{\mathrm{Set}}(X,\mathbb Z/2\mathbb Z)$ is the characteristic function. i.e. The subset of $X$$X$​​​ is one one corresponding to the characteristic function.

This lead to another equivalent definition of subobject classifier when the terminal object in $\mathcal{C}$$\mathcal C$​ exists.

Subobject classifier, the second definition.

Definition. In a category $\mathcal{C}$$\mathcal C$ with a terminal object 1 and pullbacks, an object Ω and arrow $\top :1\rightarrowtail \Omega$$⊤: 1 ↣ Ω$ provide a subobject classifier $(\Omega ,\top )$$(Ω, ⊤)$ if and only if for any $(S,s:S\rightarrowtail X)$$(S, s: S ↣ X)$ there is a unique characteristic arrow ${\chi }_s:X\to \Omega$$\chi_s : X → Ω$ making this a pullback square:

In other word, the subobject $(S,s)$$(S,s)$ is represented by ${\chi }_s:X\to \mathrm{\Omega }$$\chi_s:X\to\Omega$.

Lemma. Let $1$$1$ be a terminal object in $\mathcal{C}$$\mathcal C$, then for any object $X$$X$, $\top :1\to X$$⊤:1\to X$ is monomorphism if the morphism exists.

Proof. Let $f,g:Y\to X$$f,g:Y\to X$ such that $\top \circ f=\top \circ g$$⊤\circ f=⊤\circ g$. Since $1$$1$ is the terminal object, $f=g$$f=g$. $◻$$\square$

We already saw two ways to define subobject classifier, when it will be equivalent?

The equivalence of two definitions.

Proposition. Let $\mathcal{C}$$\mathcal C$ be a locally small category with terminal object and pullback.

Then the two definitions of subobject classifier above are equivalent.

Proof.

Form to ${\mathrm{Sub}}_{\mathcal{C}}(-)\cong {\mathrm{Hom}}_{\mathcal{C}}(-,\mathrm{\Omega })$$\mathrm{Sub}_{\mathcal C}(-)\cong\mathrm{Hom}_{\mathcal C}(-,\Omega)$ is easy.

$$\begin{array}{}\text{(8)}& {\theta }_X:{\mathrm{Sub}}_{\mathcal{C}}(X)\to {\mathrm{Hom}}_{\mathcal{C}}(X,\mathrm{\Omega })\end{array}$$

is defined by

$$\begin{array}{}\text{(9)}& (S,s)⟼{\chi }_S\end{array}$$

it is injective since ${\chi }_S$$\chi_S$ is unique, it is surjective since give you a $\chi :X\to \mathrm{\Omega }$$\chi:X\to\Omega$​​, the pullback will gives you a subobject.

The corresonding in the morphism sides follows from this diagram directly.

The subobject $(S,s^{\prime })$$(S,s')$ correspond tp $f^{\ast }(\chi )=\chi \circ f$$f^*(\chi)=\chi\circ f$.

$$\begin{array}{}\text{(10)}& \begin{array}{c}\begin{array}{ccc}{\mathrm{Sub}}_{\mathcal{C}}(X)& \stackrel{{\theta }_X}{\to }& {\mathrm{Hom}}_{\mathcal{C}}(X,\mathrm{\Omega })\\ {\mathrm{Sub}}_{\mathcal{C}}(f)\downarrow & & {\mathrm{Hom}}_{\mathcal{C}}(X,f)\downarrow & \\ {\mathrm{Sub}}_{\mathcal{C}}(Y)& \stackrel{{\theta }_Y}{\to }& {\mathrm{Hom}}_{\mathcal{C}}(Y,\mathrm{\Omega })\end{array}\end{array}\end{array}$$

Conversely suppose that ${\mathrm{Sub}}_{\mathcal{C}}(-)\cong {\mathrm{Hom}}_{\mathcal{C}}(-,\mathrm{\Omega })$$\mathrm{Sub}_{\mathcal C}(-)\cong\mathrm{Hom}_{\mathcal C}(-,\Omega)$, then we have the following diagram

There exists a subobject $\mathrm{\Omega }$$\Omega$ correspond to ${\mathrm{id}}_{\mathrm{\Omega }}$$\mathrm{id}_{\Omega}$, we call it ${\mathrm{\Omega }}_0$$\Omega_0$ and $S={\mathrm{Sub}}_{\mathcal{C}}(\varphi )({\mathrm{\Omega }}_0)$$S=\mathrm{Sub}_{\mathcal C}(\phi)(\Omega_0)$, i.e.

Since ${\theta }_X$$\theta_X$ is natural isomorphism, the $\varphi$$\phi$ here is unique.

Now we only need to check that ${\mathrm{\Omega }}_0=1$$\Omega_0=1$​ is the terminal object.

Let $X$$X$ be arbitrary object and clearly there exists some morphism via this pullback.

By the uniqueness of $\varphi :X\to \mathrm{\Omega }$$\phi:X\to\Omega$, we know that $t_0{\varphi }^{\prime }=t_0{\varphi }^{″}$$t_0\phi'=t_0\phi''$, but $t_0$$t_0$ is a monomorphism, hence ${\varphi }^{\prime }={\varphi }^{″}$$\phi'=\phi''$, hence for all $X$$X$ there exists a unique morphism from $X$$X$ to ${\mathrm{\Omega }}_0$$\Omega_0$, hence ${\mathrm{\Omega }}_0\cong 1$$\Omega_0\cong 1$. $◻$$\square$​

We will discuss more example in next section.