# Example of a NON-effective epimorphism

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?

• 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. Jul 23 '19 at 12:19

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.
• 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? Jul 23 '19 at 14:17