Topos (0): Generalized element

IntroductionGeneralized elementMonomorphism and Generalized elementOnto generalized element

Introduction

In this essay, we introduce the concept of a generalized element. Here, an element refers to the input of a function. The motivation to define a generalized element is that, in general, a morphism in a category is not a function. However, we can use generalized elements to view arbitrary morphisms as functions that map a generalized element to another generalized element. Moreover, in general, a monomorphism and an epimorphism are not isomorphisms. But with the language of generalized elements, a morphism $f$$f$ is an isomorphism if and only if $f$$f$ is one-to-one and onto on generalized elements.

Generalized element

Let $\mathcal{C}$$\mathcal C$ be a category and $x:A\to B$$x:A\to B$ be a morphism, we can think $x$$x$ as a kind of element of $B$$B$, we call that generalized element of $B$$B$ define over $A$$A$, denote as $x{\in }_{A}B$$x\in_{A} B$. We also say $A$$A$ is the stage of definition of $x$$x$​.

For a final object $1$$1$(if it exists) and arbitrary object $A$$A$, there exists only one $x{\in }_{A}1$$x\in_{A} 1$.

For any $f:B\to C$$f:B\to C$, we could write $f(x)$$f(x)$ for the composition $f\circ x$$f\circ x$. For any $x{\in }_{A}B$$x\in_{A} B$, we have $f(x){\in }_{A}C$$f(x)\in_A C$​.

That is, $f:B\to C$$f:B\to C$ will map a $A$$A$ element in $B$$B$ to a $A$$A$ element in $C$$C$. This is a well defined function from ${\mathrm{Hom}}_{\mathcal{C}}(A,B)\to {\mathrm{Hom}}_{\mathcal{C}}(A,C)$$\mathrm{Hom}_{\mathcal C}(A,B)\to\mathrm{Hom}_{\mathcal C}(A,C)$. Notice that it is nothing but pushforward $f_{\ast }$$f_*$.

If we fixed a object $A$$A$ to be the stage and think about the category of generalized elements in $\mathcal{C}$$\mathcal C$, then you get the cosmic category $A\downarrow \mathcal{C}$$A\downarrow \mathcal C$. For example, let $R$$R$ be a commutative ring, then the category of $R-$$R-$Algebra is $R\downarrow \mathrm{Ring}$$R\downarrow\mathrm{Ring}$. A $R-$$R-$Algebra $B$$B$ is just a $x{\in }_{R}B$$x\in_R B$.

Proposition. Take any $f,g:A\to B$$f,g:A\to B$, $f=g$$f=g$ if and only if for any stage $T$$T$ and every $x{\in }_{T}A,f(x)=g(x)$$x\in_{T} A,f(x)=g(x)$.

Proof. The only if is obviously. Since $f=g⟹f(x)=g(x)$$f=g\implies f(x)=g(x)$. For the if part, let $x={\mathrm{id}}_A$$x=\mathrm{id}_A$ then done. $◻$$\square$

Monomorphism and Generalized element

Let $f:B\to C$$f:B\to C$ be a monomorphism, that is, for any $g,h:A\to B$$g,h:A\to B$, $f\circ g=f\circ h⟹g=h$$f\circ g= f\circ h\implies g=h$. Now if we use the language of generalized element, we could view $f(g)=f(h){\in }_{A}C⟹g=h{\in }_{A}B$$f(g)=f(h)\in_A C\implies g=h\in_A B$. That is, a injective function from ${\mathrm{Hom}}_{\mathcal{C}}(A,B)\to {\mathrm{Hom}}_{\mathcal{C}}(A,C).$$\mathrm{Hom}_{\mathcal C}(A,B)\to\mathrm{Hom}_{\mathcal C}(A,C).$​

Onto generalized element

Similarly, we say that $f:B\to C$$f:B\to C$ is a onto on generalized element if for any $A$$A$ and $\mathrm{\forall }y:A\to C$$\forall y:A\to C$, there exists a generalized element $x{\in }_{A}B$$x\in_AB$ such that $f(x)=y$$f(x)=y$.

Proposition. $f:B\to C$$f:B\to C$ is onto on generalized element if and only if there exists a $x{\in }_{C}B$$x\in_C B$ such that $f(x)={\mathrm{id}}_C$$f(x)=\mathrm{id}_C$.

Proof. If $f$$f$ is onto, then by definition, for $y={\mathrm{id}}_C:C\to C$$y=\mathrm{id}_C:C\to C$,(here we have $A=C$$A=C$) there exists a generalized element $x{\in }_{C}B$$x\in_CB$ such that $f(x)={\mathrm{id}}_C$$f(x)=\mathrm{id}_C$. If there exists a $x{\in }_{C}B$$x\in_C B$ such that $f(x)={\mathrm{id}}_C$$f(x)=\mathrm{id}_C$, let $y{\in }_{A}C$$y\in_A C$ be a generalized element and

we have $z=x\circ y{\in }_{A}B$$z=x\circ y\in_A B$ and $f(z)=y$$f(z)=y$​.

Notice that the definition of epimorphism is:

for $g,h:Y\to Z$$g,h:Y\to Z$, we call $f:X\to Y$$f:X\to Y$ epimorphism if $g\circ f=h\circ f⟹g=h$$g\circ f=h\circ f\implies g=h$. Hence onto generalized element is epimorphism.

We say an epimorphism is split if it has right inverse. Hence $f$$f$ is a onto on generalized element iff $f$$f$ is split epimorphism.

Proposition. A morphism $f$$f$ is isomorphism iff $f$$f$ is one to one and onto on generalized element.

This is equivalent to say $f$$f$ is mono and split epi. Suppose $g$$g$ is the right inverse of $f$$f$, i.e. $f\circ g=\mathrm{id}$$f\circ g=\mathrm{id}$. Then $f\circ g\circ f=f$$f\circ g\circ f=f$.

Since $f$$f$ is mono, $f(g\circ f)=f(\mathrm{id})⟹g\circ f=\mathrm{id}.◻$$f(g\circ f)=f(\mathrm{id})\implies g\circ f=\mathrm{id}.\quad\square$