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Let $\pi: Z \rightarrow Z_2$ be the canonical epimorphism. The induced map $$ \overline{\pi}:\operatorname{Hom}(Z_2,Z)\to \operatorname{Hom}(Z_2,Z_2) $$ is the zero map. Since $\operatorname{Hom}(Z_2,Z_2) \neq 0 $ (since $\operatorname{Hom}(Z_2,Z_2) \cong Z_2 $), $\overline{\pi}$ is not an epimorphism.

This is from Hungerford book, section $4.4$:

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I don't know if it is a typo but if I follow the definition

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then:

The induced map $\overline{\pi}:\operatorname{Hom}(Z_2,Z)\to \operatorname{Hom}(Z,Z)$ is indeed the zero map since $\operatorname{Hom}(Z_2,Z)=\{0\}$ , we can say that $\operatorname{Hom}(Z,Z)\neq \{0\}$ so that $\overline{\pi}$ is not an epimorphism.

Please would you confirm if there is a typo or if I am missing something?

mrtaurho
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NotaChoice
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2 Answers2

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You have $\varphi=\pi$ going from $A=\mathbb Z$ to $C=\mathbb Z/2\mathbb Z$, hence the induced map $\overline\varphi=\overline\pi$ goes from

$$ \operatorname{Hom}(C,R)=\operatorname{Hom}(\mathbb Z/2\mathbb Z,R)\quad\text{to}\quad\operatorname{Hom}(A,R)=\operatorname{Hom}(\mathbb Z,R) $$

for some module $R$. As you can see, we cannot choose $R$ such that we get the setup from Hungerford, so this is most likely a typo. As you have observed a counterexamples arises regardless if we take $R=\mathbb Z$.

mrtaurho
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This is not a typo: you are referencing the wrong theorem. You should instead reference to Theorem 4.4, which is the COvariant version of the result you need. Unfortunately, Hungerford uses overbars for both induced maps $\overline{\phi}$. When $\phi: A \to B$, he writes $\overline{\phi}$ both for $$ \overline{\phi} : \operatorname{Hom}(C,A) \to \operatorname{Hom}(C,B) $$ as well as for $$ \overline{\phi} : \operatorname{Hom}(B,C) \to \operatorname{Hom}(A,C). $$ You need the first version. where $C=\mathbb{Z}_2$ is fixed.

Many authors use different notations for these different induced maps precisely to avoid this confusion. It is common to use $\phi_*$ for the covariant version and $\phi^*$ for the contravariant.

Randall
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  • Thanks for the answer, so the covariant map is when the map is between $Hom(X,R)$ and contravariant when between $Hom(R,X)$ ? which is which – NotaChoice Dec 13 '21 at 21:59
  • Covariant is first slot fixed, second slot variable. Contravariant is the vice versa. – Randall Dec 13 '21 at 22:05
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    @Omar Conventionally, to make it clear which place is the "blank to fill in" (after all, $X$ could also be some established, fixed object; letters are treacherous like that), we use $-$ or ${}\bullet{}$. So it's the covariant $Hom(A,-)$ (or $Hom(A,{}\bullet{})$) and contravariant $Hom(-, B)$. – Arthur Dec 13 '21 at 22:11
  • thanks that is good to know, any idea where the names covariant and contravariant come from? cause I m sure I m gonna forget that soon if I can't associate it to anything – NotaChoice Dec 13 '21 at 23:28
  • See chapter 10, section 1 of Hungerford. The language is from category theory. @Omar – Randall Dec 14 '21 at 01:21