이미지(범주 이론)

Let ${\displaystyle \alpha ,\,\beta }$ be such that ${\displaystyle \alpha \,e=\beta \,e}$, one needs to show that ${\displaystyle \alpha =\beta }$. Since the equalizer of ${\displaystyle (\alpha ,\beta )}$ exists, ${\displaystyle e}$ factorizes as ${\displaystyle$$$$q$ ${\$displaystyle $q}$ monic을$$ $사용$하는 e$=q\,e'}$그러나 $f{\displaystyle }$=${\displaystyle }$( ${\displaystyle m}$ ${\displaystyle q{}^{}}$) e ${\displaystyle {\prime }^{}}$ ${\$ f$=(m\,q)\,e'}$은$({\displaystyle m}$ ${\displaystyle q}$) ${\displaystyle$f$}$의 $단일형성$을 $갖는$f ${\$$displaystyle$f}의 인자화다$.$따라서 영상의 범용 특성에 의해 ${\displaystyle \mathrm{고유}}$한 화살표 $v{\displaystyle }$: I${\displaystyle }$→ ${\displaystyle E}$ ${\displaystyle {}_{}}$ ${\displaystyle {\alpha }_{}}$, ${\displaystyle {\beta }_{}}$ ${\$$I\to Eq_{\alpha ,\$$beta }$ 이렇게$$${\displaystyle }$ m${\displaystyle }$= ${\displaystyle m}$ ${\displaystyle q}$ ${\displaystyle v}$ ${\$$displaystyle m}$이(가$)$ 모닉 ${\displaystyle {\text{ID}}_{}}$ ${\displaystyle {}_{}}$=$$ v {\$displaystyle{\textid}}}{{\$displaystysty}}{\$text}}}}}{{}}}}}}}}}}}}}}}}}}}}$$I}=q\,v}$. Furthermore, one has ${\displaystyle m\,q=(mqv)\,q}$ and by the monomorphism property of ${\displaystyle mq}$ one obtains ${\displaystyle {\text{id}}_{Eq_{\alpha ,\beta }}=v\,q}$.

이것은 $I{\displaystyle }$ ${\displaystyle \equiv }$ ${\displaystyle E}$ ${\displaystyle {}_{}}$ ${\displaystyle {\alpha }_{}}$ ,${\displaystyle {\beta }_{}}$ ${\$을 의미하며, 따라서$$ ${\displaystyle {\text{ID}}_{}}$ ${\displaystyle {}_{}}$= ${\displaystyle q}$ ${\displaystyle v}$ ${\displaystyle{\text{id}}}_{{}}}}}}}}{{}}}}}}}}}}}}$$I}=q\,v}$은$({\displaystyle \alpha }$ ,${\displaystyle \beta }$) ${\displaystyle (\alpha ,\beta$ $)\},$ whence $\alpha {\displaystyle }$= ${\displaystyle \beta }$ ${\displaystyle \alpha =\beta$$}.$

첫 번째 정의는 두 번째 정의를 의미한다.(1)이 $m{\displaystyle }$ ${\displaystyle m}$ 정규$$ 단형성을 유지한다고 가정한다.

• Equalization: one needs to show that ${\displaystyle i_{1}\,m=i_{2}\,m}$ . As the cokernel pair of ${\displaystyle f,\ i_{1}\,f=i_{2}\,f}$ and by previous proposition, since ${\displaystyle C}$ has all equalizers, the arrow ${\displaystyle e}$ in the factorization $$${\displaystyle f}$= ${\displaystyle m}$ ${\displaystyle e}$ ${\displaystyle f=m\,e}$은(는) 경구형이기 때문에 i ${\displaystyle 1{}_{}}$ ${\displaystyle {f\mathrm{}}_{}}$= i ${\displaystyle {2\mathrm{}}_{}}$ ${\displaystyle f{}_{}}$ ${\displaystyle \text{⇒}}$ i = ${\displaystyle \text{i}{}_{}}$ = ${\displaystyle 1m{}_{}}$= ${\displaystyle {}_{}}$ ${\displaystyle 2m{}_{}}$ ${\displaystyle i_{1}\,f=i_{2}\\rightarrow$\$i_{1}\,m=i_{2$},$m$
• 보편성: 모든 콜리미트(또는 적어도 모든 푸시아웃)가 있는 범주에서 ${\displaystyle m}$ 자체는$$ 코커넬 쌍$({\displaystyle Y}$ ${\displaystyle {}_{}}$ ${\displaystyle I_{}}$ ${\displaystyle {}_{}}$, ${\displaystyle {}_{}}$ ${\displaystyle {1\mathrm{}}_{}}$, ${\displaystyle {c\mathrm{}}_{}}$ ${\displaystyle {}_{}}$) ${\displaystyle(Y\sqcup _{I}Y_c_{1},c_{2}}}}}$을 허용한다.

더욱이, 일반적인 단형주의로서$,{\displaystyle }$ (${\displaystyle I}$,${\displaystyle }$m ) ${\displaystyle (I,$m)}은$$ 한 쌍의 $형태론{\displaystyle {}_{}{b\mathrm{}}_{}}$ 1, ${\displaystyle {b\mathrm{}}_{}}$ : ${\displaystyle Y}$⟶ ${\displaystyle B}${\$displaystyle b_{1},b_{2}$의 등분자이다.$Y\longwrightarrow B}$ 그러나$$ 여기서 우리는 그것이 $또한{\displaystyle {}_{}}$ ${\displaystyle c}$ ${\displaystyle {1\mathrm{}}_{}}$,${\displaystyle {}_{}{c\mathrm{}}_{}}$ 2${\displaystyle {}_{}}$: ${\displaystyle Y}$ ${\displaystyle ⟶}$ Y ⊔ ${\displaystyle I_{}}$ ${\displaystyle {}_{}}$ ${\displaystyle {}_{}}$ ${\displaystyle c_{1},c_{2$}의$$등가제라고 주장한다.$Y\longwrightarrow$ Y$\sqcup _{I}Y$

$실제로{\displaystyle }$, 시공 $b{\displaystyle {}_{}}$ ${\displaystyle 1m{}_{}}$= ${\displaystyle {}_{}}$ ${\displaystyle 2m{}_{}}$ ${\displaystyle b_{1}\,m=b_{2}\,m}$ 따라서$$ m ${\displaystyle$ m$}$에 대한 "코커넬 쌍" 다이어그램은 고유한 형태론 $u{\displaystyle {}^{}}$u${\displaystyle }$: ${\displaystyle Y{}_{}⟶}$ ${\displaystyle B}$ ${\displaysty u'$를 산출한다$$.$Y\sqcup _{I}$$B$$$${\displaystyle {1\mathrm{}}_{}}$= ${\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ ${\displaystyle c{}_{}}$ ${\displaystyle {1\mathrm{}}_{}}$,${\displaystyle {}_{}\text{b}{}_{}}$ ${\displaystyle {2\mathrm{}}_{}}$= ${\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ c ${\displaystyle {2\mathrm{}}_{}}$ ${\$ 이제 지도 $m{\displaystyle {}^{}}$′${\displaystyle }$: ${\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}Y}$ ${\displaystyle m':$$I'\longrightarrow Y}$ which equalizes ${\displaystyle (c_{1},c_{2})}$ also satisfies ${\displaystyle b_{1}\,m'=u'\,c_{1}\,m'=u'\,c_{2}\,m'=b_{2}\,m'}$, hence by the equalizer diagram for ${\displaystyle (b_{$$1,b_{2$ 고유 맵 $h{\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ :${\displaystyle I}$ ${\displaystyle {\prime }^{}}$→ I ${\displaystyle h':$$그렇게$ m$$${\displaystyle {\prime }^{}}$= ${\displaystyle m}$ ${\displaystyle h{}^{}}$ ${\displaystyle {\prime }^{}}$ ${\$ m$'=m\,h$

마지막으로, j ${\displaystyle {1\mathrm{}}_{}}$ ${\displaystyle :=}$ ${\displaystyle {c1\mathrm{}}_{}}$,${\displaystyle {}_{}{j\mathrm{}}_{}}$ 2 ${\displaystyle :=}$ c ${\displaystyle {2\mathrm{}}_{}}$,${\displaystyle {}_{}\text{Z}}$ : ${\displaystyle :=}$ ${\displaystyle Y{}_i}$ I ${\displaystyle {}_{}}$ ${\displaystyle j_{1}:=c_{1$},$j_{2}:=c_{2}, z:$와 함께 cokernel 쌍 다이어그램($f{\displaystyle }${\$displaysty f$을 사용하십시오.$=Y\sqcup _{I}Y}$ : 고유한 $u{\displaystyle :}$ ${\displaystyle Y{}_x}$ ${\displaystyle {}_{}}$ Y ${\displaystyle ⟶}$ ${\displaystyle Y}$ ${\displaystyle {}_{}}$ ${\displaystyle I_{}}$ ${\displaystyle {}_{}}$ ${\displaystyle u:$$Y\sqcup _{X}Y\longrightarrow Y\sqcup _{I}Y}$ such that ${\displaystyle c_{1}=u\,i_{1},\ c_{2}=u\,i_{2}}$. Therefore, any map ${\displaystyle g}$ which equalizes ${\displaystyle (i_{1},i_{2})}$ also equalizes ${\displaystyle (c_{1},c_{$${}}$을(를)${\displaystyle }$g = m $\$ {\$displaystyle g=m$\,h$'}$으로 고유하게 인수함$.$ 이는 정확히${\displaystyle }$( ${\displaystyle {}_{}}$${\displaystyle ,}$ m${\displaystyle }$) ${\displaystyle(I, m)$이${\displaystyle }$( i ${\displaystyle {1\mathrm{}}_{}}$,${\displaystyle {}_{}{i\mathrm{}}_{}}$ ${\displaystyle {}_{}}$) ${\displaystyle(i_{1},i_{2})$의 동일자임을 의미한다$.$

두 번째 정의는 첫 번째 정의를 의미한다.

• 인자화: 이퀄라이저 다이어그램에서$$ $m{\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ ${\displaystyle :=f}$ ${\displaystyle$ m$'}$을$($를) 취하면($$${\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ ${\displaystyle g}$에 해당$$), $인수화{\displaystyle }$ f${\displaystyle }$= ${\displaystyle m}$ ${\displaystyle f=m\,h}$을(를) 얻는다$.$
• 보편성: let $f{\displaystyle }$= ${\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ ${\displaystyle e{}^{}}$′ ${\$은(는) $m{\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ ${\$의 정규 단모형, 즉, 일부 쌍${\displaystyle {}_{}{d\mathrm{}}_{}}$ ${\displaystyle {}_{}}$, d${\displaystyle {}_{}{2\mathrm{}}_{}}$ )의 등가형제$(d_{1},d_{2$

Then ${\displaystyle d_{1}\,m'=d_{2}\,m'\ \Rightarrow \ d_{1}\,f=d_{1}\,m'\,e=d_{2}\,m'\,e=d_{2}\,f}$ so that by the "cokernel pair" diagram (of ${\displaystyle f}$), with ${\$$displaystyle j_{1}:=d_{1},\ j_{2}:=d_{2},\ Z:$$=D$ 고유한 $u{\displaystyle {}^{}″:}$ ${\displaystyle Y}$ ${\displaystyle {}_{}}$ ${\displaystyle X_{}}$ ${\displaystyle {}_{}}$ ${\displaystyle ⟶}$ ${\displaystyle D}$ ${\displaystyle u'$가 있음:${\displaystyle Y_{}}$$\sq컵 _{X}Y\longwrightarrow D},$ d 1${\displaystyle {}_{}}$= ${\displaystyle {}^{}}$u i ${\displaystyle {1\mathrm{}}_{}}$,${\displaystyle {}_{}}$d ${\displaystyle {2\mathrm{}}_{}}$= ${\displaystyle {}^{}}$ ${\displaystyle {″}^{}}$ i ${\displaystyle {2\mathrm{}}_{}}$ ${\$2$\}\}.$

이제,에서 나는 1m)나는 2m{\displaystyle i_{1}\,m=i_{2}\,m}(i1(의 이퀄라이저, i2에서 m)도표), 하나 얻d1m)u″ 나는 1m)u″ 나는 2m)d2m{\displaystyle d_{1}\,m=u"\,i_{1}\,m=u"\,i_{2}\,m=d_{2}\,m}, 따라서의 보편성의(1(의 이퀄라이저, d2), 다이어그램과 f에.m),${\displaystyle \mathrm{고유}}$한 ${\displaystyle v:}$I ${\displaystyle {}^{}}$ ${\displaystyle {\prime }^{}}$ I$display$ {\$displaystyle$ v$:Im$\longrightarrow $I'}$이(가) 있으며, $이러한$ v = ${\displaystyle {}^{}}$ = m${\displaystyle {}^{}m}$ v$${\$displaystyle m=m'\,$v}이$($가) 있다$.$