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monomorphism [2015/03/16 23:44]
nikolaj
monomorphism [2015/03/17 14:08]
nikolaj
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 $f(x)=g(y)\implies x=y$. $f(x)=g(y)\implies x=y$.
 +
 +=== Theorems ===
 +The pullback of a mono is mono.
  
 === Discussion === === Discussion ===
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 Say $f$ is not an injection and hence collapses two different terms $a,d\in A$ into a single value, i.e. $f(a)=f(d)$. The pullback object then contains $\langle a,a\rangle$ and $\langle d,​d\rangle$,​ but also $\langle a,d\rangle$ and $\langle d,​a\rangle$. So $A\times_BA$ is bigger than $A$. On the other hand, if $f$ is an injection, then for any $a$, the pullback only contains $\langle a,​a\rangle$. We get $A\times_BA\cong A$.  Say $f$ is not an injection and hence collapses two different terms $a,d\in A$ into a single value, i.e. $f(a)=f(d)$. The pullback object then contains $\langle a,a\rangle$ and $\langle d,​d\rangle$,​ but also $\langle a,d\rangle$ and $\langle d,​a\rangle$. So $A\times_BA$ is bigger than $A$. On the other hand, if $f$ is an injection, then for any $a$, the pullback only contains $\langle a,​a\rangle$. We get $A\times_BA\cong A$. 
  
-The definition "​$\langle A,​\prod_{A}1_A\rangle$ is a pullback of $f$ along itself"​ implies that $A$ is already a valid pullback object, because $f$ doesn'​t collapse any information.+The definition "​$\langle A,​\prod_{A}1_A\rangle$ is a pullback of $f$ along itself"​ implies that $A$ is already a valid pullback object, because $f$ doesn'​t collapse any information. Dually, the definition of an epimorphism (surjections in ${\bf{Set}}$) implies that $A$ is a valid pushout $A+_BA$, an object which generally contains less information than $A$.
  
 A stepwise characterization of a monomorphism from the pullback, with a focus on the universal property, is given in [[pullback . category theory|pullback]]. A stepwise characterization of a monomorphism from the pullback, with a focus on the universal property, is given in [[pullback . category theory|pullback]].
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