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monomorphism [2015/03/16 23:44] nikolaj |
monomorphism [2015/03/17 14:08] nikolaj |
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$f(x)=g(y)\implies x=y$. | $f(x)=g(y)\implies x=y$. | ||
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+ | === 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]]. |