Topos (1): Subobject

MotivationNotationSubobject,monomorphism

 

Motivation

In $\mathrm{Set}$$\mathrm{Set}$, we can talk about subset of another set, denote as $S\subseteq X$$S\subseteq X$. But this point of view is unnatural when you think about another category such that $\mathrm{Mod}(R),\mathrm{Top},$$\mathrm{Mod}(R),\mathrm{Top},$ or some non-concrete category. Let $M$$M$ be a $R-$$R-$module, a subset of $M$$M$ in general is not a submodule of $M$$M$. Let $X$$X$ be a topological space, a subset of $X$$X$ is not a subspace of $X$$X$. Let $\mathcal{N}$$\mathcal N$ be a category whose object are natural number and ${\mathrm{Hom}}_{\mathcal{N}}(n,m)=M_{m\times n}(\mathbb{F})$$\mathrm{Hom}_{\mathcal N}(n,m)= M_{m\times n}(\mathbb F)$, here $\mathbb{F}$$\mathbb F$ is a field. Then easy to see $\mathcal{N}\cong \mathrm{FinVect}(\mathbb{F})$$\mathcal N\cong\mathrm{FinVect}(\mathbb F)$. But you could not think about the subobject of $n\in \mathrm{Ob}(\mathcal{N})$$n\in\mathrm{Ob}(\mathcal N)$ via the set theoretical language.

Set theory is not the correct language when you think about some $\mathcal{C}\ne \mathrm{Set}$$\mathcal C\ne\mathrm{Set}$. We try to think element freely via category theory.

Think about a subset $S\subseteq X$$S\subseteq X$. We know we could replace the $\subseteq$$\subseteq$ by the inclusion map $i:S\hookrightarrow X$$i:S\hookrightarrow X$, that is, a monomorphism. Let us generalize this idea.

Notation

In this essay, $\hookrightarrow$$\hookrightarrow$ always means monomorphism. We will abbreviate monomorphism to mono, and $f\circ g$$f\circ g$ as $fg$$fg$.

Denote ${\mathcal{C}}^{\prime }$$\mathcal C'$ as the subcategory such that $\mathrm{Ob}({\mathcal{C}}^{\prime })=\mathrm{Ob}(\mathcal{C})$$\mathrm{Ob}(\mathcal C')=\mathrm{Ob}(\mathcal C)$ but ${\mathrm{Hom}}_{{\mathcal{C}}^{\prime }}(X,Y)$$\mathrm{Hom}_{\mathcal C'}(X,Y)$ be the monomorphism from $X$$X$ to $Y$$Y$.

Subobject,monomorphism

Definition. Let $\mathcal{C}$$\mathcal C$ be a category, $X$$X$ be a object in $\mathcal{C}$$\mathcal C$, we say a subobject of $X$$X$ is a pair $(S,s)$$(S,s)$ such that

$$\begin{array}{}\text{(1)}& S\stackrel{s}{\hookrightarrow }X\end{array}$$

We could view $S$$S$ as a part of $X$$X$ via the mono $s$$s$.

Definition. Let $(S,s),(R,r)$$(S,s),(R,r)$ be two subobject, we define $(R,r)\subseteq (S,s)$$(R,r)\subseteq (S,s)$ (that is the morphism between subobjects) iff there is a morphism $h:R\to S$$h:R\to S$ such that

$$\begin{array}{}\text{(2)}& r=sh\end{array}$$

Well, would we need to require $h$$h$ to be mono? Indeed, no, since $h$$h$ has to be mono by next proposition.

Proposition. Let $s:A\to B,g:B\to C$$s:A\to B,g:B\to C$ be two morphism. $gs$$gs$ is mono implies $s$$s$ is mono.

Proof. Let $h,k$$h,k$ be two morphism such that $sh=sk$$sh=sk$. Then $g(sh)=g(sk)⟹(gs)h=(gs)k$$g(sh)=g(sk)\implies (gs)h=(gs)k$. Since $gs$$gs$ is mono, we get that $h=k$$h=k$. Hence $sh=sk⟹h=k$$sh=sk\implies h=k$, $s$$s$ is mono, and the morphism between two subobject is unique. $◻$$\square$

Hence the category of subobjects of $X$$X$ is the slice category ${\mathcal{C}}^{\prime }/X$$\mathcal C'/X$, and indeed, this is a preorder set.

Proposition. If $(R,r)\subseteq (S,s)$$(R,r)\subseteq (S,s)$ and $(S,s)\subseteq (R,r)$$(S,s)\subseteq (R,r)$, then $R\cong S$$R\cong S$.

Proof. Let $(R,r)\stackrel{h}{\hookrightarrow }(S,s)$$(R,r)\xhookrightarrow{h} (S,s)$ and $(R,r)\stackrel{h^{\prime }}{\hookleftarrow }(S,s)$$(R,r)\xhookleftarrow{h'} (S,s)$ be two mono in ${\mathcal{C}}^{\prime }/X$$\mathcal C'/X$. Then $h{h}^{\prime }$$hh'$ is the unique morphism from $(S,s)$$(S,s)$ to itself, i.e. ${\mathrm{id}}_S$$\mathrm{id}_{S}$. Similarly for $h^{\prime }h.$$h'h.$ $◻$$\square$

Hence we define subobject only up to isomorphism. Let us back to $\mathrm{Set}$$\mathrm{Set}$ and see what happens.

Example. Let us consider two subobjects of $\mathbb{Z}$$\mathbb Z$ in $\mathrm{Set}$$\mathrm{Set}$. $(2\mathbb{Z},i)$$(2\mathbb Z,i)$ and $(\mathbb{Z},2)$$(\mathbb Z,2)$. Here $i$$i$ is the inclusion map and $2$$2$ is $x\mapsto 2x$$x\mapsto 2x$.

We claim that $(2\mathbb{Z},i)\cong (\mathbb{Z},2)$$(2\mathbb Z,i)\cong (\mathbb Z,2)$. Consider $(\mathbb{Z},2)\stackrel{2}{\hookrightarrow }(2\mathbb{Z},i)$$(\mathbb Z,2)\xhookrightarrow{2}(2\mathbb Z,i)$. Easy to see that $\mathbb{Z}\stackrel{i2}{\hookrightarrow }\mathbb{Z}=\mathbb{Z}\stackrel{2}{\hookrightarrow }\mathbb{Z}$$\mathbb Z\xhookrightarrow{i2}\mathbb Z=\mathbb Z\xhookrightarrow{2}\mathbb Z$. Hence $(\mathbb{Z},2)\subseteq (2\mathbb{Z},i)$$(\mathbb Z,2)\subseteq (2\mathbb Z,i)$.

Now consider $(2\mathbb{Z},i)\stackrel{\frac{1}{2}}{\hookrightarrow }(\mathbb{Z},2)$$(2\mathbb Z,i)\xhookrightarrow{\frac{1}{2}}(\mathbb Z,2)$. Easy to see that $2\mathbb{Z}\stackrel{2\frac{1}{2}}{\hookrightarrow }\mathbb{Z}=2\mathbb{Z}\stackrel{i}{\hookrightarrow }\mathbb{Z}$$2\mathbb Z\xhookrightarrow{2\frac{1}{2}}\mathbb Z=2\mathbb Z\xhookrightarrow{i}\mathbb Z$. Hence $(\mathbb{Z},2)\cong (2\mathbb{Z},i)$$(\mathbb Z,2)\cong(2\mathbb Z,i)$.

This is not a coincidence, indeed, we have the following proposition.

Proposition. In $\mathrm{Set},\mathrm{Grp},\mathrm{Ring},\mathrm{Mod}(R)$$\mathrm{Set,Grp,Ring,}\mathrm{Mod}(R)$, $(R,r)\cong (S,s)$$(R,r)\cong (S,s)$ if and only if $\mathrm{Im}(r)=\mathrm{Im}(s)$$\mathrm{Im}(r)=\mathrm{Im}(s)$.

Proof. Let $(R,r),(S,s)$$(R,r),(S,s)$ be two subobjects and $(R,r)\cong (S,s)$$(R,r)\cong (S,s)$. By definition $(R,r)\subseteq (S,s)$$(R,r)\subseteq (S,s)$ implies that $r=sh$$r=sh$.

Hence $\mathrm{Im}(r)\subseteq \mathrm{Im}(s)$$\mathrm{Im}(r)\subseteq\mathrm{Im}(s)$. Hence $(R,r)\cong (S,s)⟹\mathrm{Im}(r)=\mathrm{Im}(s)$$(R,r)\cong (S,s)\implies\mathrm{Im}(r)=\mathrm{Im}(s)$.

Conversely, let $\mathrm{Im}(r)=\mathrm{Im}(s)$$\mathrm{Im}(r)=\mathrm{Im}(s)$. Then consider $\mathrm{Im}(r)\stackrel{r^{-1}}{\hookrightarrow }R$$\mathrm{Im}(r)\xhookrightarrow{r^{-1}} R$ and $s=r(r^{-1}s)$$s=r(r^{-1}s)$, hence $(S,s)\subseteq (R,r)$$(S,s)\subseteq (R,r)$. Similarly, $(R,r)\subseteq (S,s)$$(R,r)\subseteq (S,s)$. Hence $(R,r)\cong (S,s)$$(R,r)\cong(S,s)$. $◻$$\square$

So, let us go back to the category $\mathcal{N}$$\mathcal N$, a subobject of $n$$n$ is $(m,M_{n\times m})$$(m,M_{n\times m})$, where $M_{n\times n}$$M_{n\times n}$ is a matrix. Also, we know that $M_{n\times m}$$M_{n\times m}$ is mono implies $m\le n$$m\le n$.

Well, what is the connection between subspace topology of $(X,\tau )$$(X,\tau)$ and subobject of $(X,\tau )$$(X,\tau)$?

Let $((Y,{\tau }_j),i)$$((Y,\tau_j),i)$ be a subobject of $(X,\tau )$$(X,\tau)$. Fix the subset $Y$$Y$ and the inclusion map $i$$i$. Consider the subcategory of ${\mathrm{Top}}^{\prime }/X$$\mathrm{Top}'/X$ whose objects are $(Y,{\tau }_j),j\in J$$(Y,\tau_j),j\in J$. Then the subspace topology is the initial object of this subcategory.