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I'm reading Introduction to Étale Cohomology by Tamme and I'm confused by the notion of effective epimorphism (page 25, section 1.3.1).

Recall that an epimorphism (in a category $\mathbf{C}$) is a morphism $U\overset{\phi}\rightarrow V$ such that the induced morphism of set $\operatorname{Hom}(V,T)\overset{\phi^o}\rightarrow\operatorname{Hom}(U,T)$ is injective for any object $T$ in $\mathbf{C}$.

Suppose that the fiber product $U\times_V U$ exists.

Then we say that $U\rightarrow V$ is an effective epimorphism if, in the diagram $$\operatorname{Hom}(V,T)\overset{\phi^o}\rightarrow\operatorname{Hom}(U,T)\overset{\pi_1^o}{\underset{\pi_2^o}\rightrightarrows}\operatorname{Hom}(U\times_V U,T),$$ $\phi^o$ is an equalizer. Here $\pi_1^o$ and $\pi_2^o$ denote the precomposition with the projections of the first and second coordinate of $U \times_V U$.

My question: I have the funny wrong feeling that every epimorphism is effective, because I feel that every epimorphism equalize the diagram above. Can someone give me an example of an epimorphism that is not effecttive? Isn't it always true that for every $f\in \operatorname{Hom}(V,T)$, the maps $f\circ\phi\circ\pi_1,f\circ\phi\circ\pi_2\in\operatorname{Hom}(U\times_V U,T)$ are the same?

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  • $\begingroup$ I think you have some confusion over what it means for $\phi^0$ to be an equalizer. It doesn't just mean that $\pi_1^0 \circ \phi^0=\pi_2^0 \circ \phi^0$, but it also has to satisfy the universal property of an equalizer. $\endgroup$ Jul 23 '19 at 12:19
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Consider the category of rings $\mathbf{Ring}$ and the epimorphism $i:\Bbb Z \to \Bbb Q$. Then I claim that $i$ is not regular. Indeed, we have $\Bbb Z \times_{\Bbb Q} \Bbb Z=\{(a,b) \in \Bbb Z^2\mid i(a)=i(b)\}$ As $i$ is injective, this is just the diagonal inside $\Bbb Z^2$, so $\pi_1,\pi_2$ are isomorphisms and we even have $\pi_1=\pi_2$. This implies that $\pi_1^0=\pi_2^0$ for any object $T$. Since an equalizer of two equal arrows is an isomorphism, if we assume that $i$ is an effective epimorphism, we get that $i^0:\mathrm{Hom}(V,T) \cong \mathrm{Hom}(U,T)$ is a natural bijection, thus $i$ is an isomorphism by the Yoneda lemma. But evidently $i$ is not an isomorphism.

The argument generalizes to show that any monomorphism which is also an epimorphism, but not an isomorphism is not an effective epimorphism.

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  • $\begingroup$ Thank you! Tamme didn't introduce the notion of regular, but now that I saw it, your example makes perfect sense. $i$ is not regular, hence not effective. Although, how can I see the difference between these two notions? Do you know an example of an epimorphism which is regular but not effective? Are these notions actually different? $\endgroup$
    – Pippo
    Jul 23 '19 at 14:17
  • $\begingroup$ Ok, one can prove that in any category with pullbacks, an epimorphism is regular iff effective. $\endgroup$
    – Pippo
    Jul 23 '19 at 14:20

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