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the definition of monomorphism as follows: enter image description here

Suppose there is a very small category, which contains only two objects X and Y, where X and Y are both sets. There are only 3 morphisms in this category: idX, idY and f: X->Y. I think it should be legal to construct a category like this.

From the set point of view, the mapping f form X to Y is not necessarily an injective. But from the perspective of category, since the morphism pointing to X is only idX, then f g1 = f g2 can lead to g1=g2. Because both g1 and g2 can only be idX. In other words, f is a monomorphism.

Sorry for the very basic question. Is my understanding of the category accurate?

Acuzio Leigh
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    The category doesn't "know" what the sets $X$ and $Y$ are, or what $f$ is as a function. So the fact that $f$ is monic doesn't tell you anything about injectivity. – Alex Provost Dec 27 '20 at 05:55
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    There's no reason why $f$ even has to be a function at all. Anyway, more generally, there are plenty of "naturally occurring" examples of concrete categories (i.e., the objects are sets with some extra structure/properties, and the morphisms are structure-preserving functions) with monomorphisms that aren't injective and epimorphisms that aren't surjective. – Daniel Hast Dec 27 '20 at 05:58
  • thanks for the reply. I constructed this example just to confirm that my understanding of the basic concepts is correct. Does this mean that although the monomorphism is very similar to the injective in the set. However, for the category constituted by sets and mapping between sets, f is monomorphism does not cause f to be injective? – Acuzio Leigh Dec 27 '20 at 06:00
  • @AcuzioLeigh: Correct; even in concrete categories, monomorphisms need not be injective functions of the underlying sets. Morphisms that are injective on underlying sets are necessarily monomorphisms, but the converse does not hold in general. See comments here and here for an example of a monomorphism that is not injective. – Arturo Magidin Dec 27 '20 at 06:03
  • The Title is baffling. What is "the injective in set theory"? Your example has one non-identity map that is noted to be not-necessarily-injective (contrary to the apparent assumption in the title)? You mention set theory in the title -- are you working in a concrete category? Why are you comparing/contrasting injective functions (which definition depends on the details of the maps on the elements of the objects, which elements are largely invisible to category theory) and monomorphisms (which definition depends on the details of maps composed with maps)? – Eric Towers Dec 27 '20 at 06:08
  • @EricTowers give the OP a break - it's obviously a beginner question, and most likely they haven't learned about non-concrete categories yet. To the OP: it's not a bad question - the answer is yes, you got it right. – N. Virgo Dec 27 '20 at 13:25

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