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Following is the definition of strict epimorphism from the paper I'm reading.

enter image description here

However, I have some confusion. What does $$ \mathcal{C}\xrightarrow{x}X $$

mean? I think $\mathcal{C}$ is a category which contains $X$ as an object, so what is $x$?

And after this definition the author says that strict epimorphism + monomorphism = isomorphism. Could anyone provide me a proof?

I'm new to category theory, forgive me if the question is stupid.

By the way, I have googled the term strict epimorphism, seeing that there is few result. Even Wiki does not have this term. Is it an isolated term?

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Shouldn't it be $g$ compatible if for all $x,y$, if $fx=fy$ then $gx=gy$? – Stefan Hamcke Oct 30 '13 at 16:27
Otherwise any monic $g$ would be compatible. – Stefan Hamcke Oct 30 '13 at 16:33
That would also be consistent with the nLab-definition. – Stefan Hamcke Oct 30 '13 at 16:39
This is probably the case; I actually didn't understand compatible at first, i thought it was a condition defined previously in the article (here) but it just means composable.; Then in the nLab definition, i don't really understand all that coequalizer business, but at least it refers to a precise book, Category and Sheaves by Kashiwara and Schapira, with a general and precise definition: 5.1.5 (e) p.115; with more search, one can also find a similar def. 8.10 p.59 in Theory of categories by Nicolae and Liliana Popescu without a priori unicity of – user39158 Apr 28 at 9:53
the factorization (Careful: 8.10 def. of strict monomorphism, not epi) – user39158 Apr 28 at 9:54
up vote 1 down vote accepted

One has to read $$C \xrightarrow {x} X {\rm \ for \ any \ object\ } C\in \mathcal{C}$$ instead of $$\mathcal{C}\xrightarrow {x} X.$$

A proof that "strict epimorphism + monomorphism = isomorphism" is in the book

I. Bucur, A. Deleanu, Introduction to the Theory of Categories and Functors, 1970, Lemma 3.17.

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Is it a typo, or a standard abbreviation? – hxhxhx88 Oct 30 '13 at 17:07
I think, it is a typo. – Boris Novikov Oct 30 '13 at 17:10
OK, thanks very much! – hxhxhx88 Oct 30 '13 at 17:11
You are welcome! – Boris Novikov Oct 30 '13 at 17:12

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