Is it possible to describe the property "being a monomorphism" as a universal property (with appropriate category/ies and functor)?


1 Answer 1


Well, it really depends on what you mean by "described as a universal property". Maybe you find this description acceptable:

  • A morphism $f : A \to B$ is a monomorphism if and only if the diagram $$\begin{array}{rcl} A & \overset{\textrm{id}}{\rightarrow} & A \\ {\scriptstyle \textrm{id}} \downarrow & & \downarrow {\scriptstyle f} \\ A & \underset{f}{\rightarrow} & B \end{array}$$ is a pullback square.

    Indeed, suppose $g, h : X \to A$ are morphisms such that $f \circ g = f \circ h$; if the diagram is a pullback, then there is a unique $k : X \to A$ such that $g = h = k$, and so $f$ is a monomorphism; and if $f$ is a monomorphism then $g = h$ so there indeed a unique morphism completing the obvious diagram.

This description is certainly useful. For example, it implies:

  • Any functor that preserves pullbacks (or even just kernel pairs) also preserves monomorphisms. In particular, right adjoints preserve monomorphisms.

  • Any functor that reflects pullbacks (or kernel pairs) and isomorphisms must also reflect monomorphisms. In particular, monadic functors reflect monomorphisms.

Or perhaps you would prefer something in terms of hom-sets:

  • A morphism $f : A \to B$ in a (locally small) category is a monomorphism if and only if $f_* : \textrm{Hom}(X, A) \to \textrm{Hom}(X, B)$ is injective for all $X$.

If you think about it, this is just the definition of ‘monomorphism’. Amusingly this can be derived in a roundabout way by noting that $\textrm{Hom}(X, -)$ is a functor that preserves pullbacks, and the collection of all such functors jointly reflects pullbacks and isomorphisms.

  • $\begingroup$ Thank you Zhen Lin, I was aware of these descriptions, but I was hoping to get something along the lines of the general definition of universal arrow (as initial/terminal object in an appropriate slice category, given an appropriate functor). $\endgroup$
    – magma
    Dec 29, 2012 at 18:30
  • $\begingroup$ There's no such definition that isn't tautological. For example, I could say that $f$ is a monomorphism if and only if it is equal to its own image... but the image of a morphism is defined to be the smallest monomorphism through which it factors. $\endgroup$
    – Zhen Lin
    Dec 30, 2012 at 0:27

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