# Class of all objects and all monomorphisms of an arbitrary category $\mathcal {C}$forms a subcategory

I am unable to convince myself the what is said in the title is true. I found it https://www.encyclopediaofmath.org/index.php/Monomorphism here. Where they have given the definition of monomorphism.

Let me explain my confusion; Suppose I take $$\mathcal{C}$$ to the the category of abelian groups . The title states that if I construct a subcategory $$\mathcal{S}$$ of $$\mathcal{C}$$ with objects as all abelian groups and morphisms as only injective group homomorphisms then $$\mathcal{S}$$ is a subcategory.

But subcategory is itself a category that means any two objects in the subcategory must have a morphism between them, but does taking only injective gp homomorphisms serves the purpose? What I mean is any two ablian groups might not have any injective map between them at all. Then how are we considering them as a subcategory?

Any hint or help is much appreciated. Thanks in advance.

• All this means is that a composite of two injective homomorphisms is an injective homomorphism. – Angina Seng Oct 24 '19 at 6:41
• "But subcategory is itself a category that means any two objects in the subcategory must have a morphism between them" Why would it mean that? – Eric Wofsey Oct 24 '19 at 6:46
• The class of morphisms between two different objects is also allowed to be empty. You only demand the existence of at least one endomorphism for every object. – Con Oct 24 '19 at 6:51
• In $\mathbf{Set}$ there are no morphisms $A\to\emptyset$ when $A$ is nonempty. – Oscar Cunningham Oct 24 '19 at 7:03
• A category definitely does not need to have an arrow between any two objects: see this question (Oscar already gave an example where there is no arrow in one direction, but there is of course still an arrow in the other direction). – Mark Kamsma Oct 24 '19 at 10:47