In category theory, a monomorphism (also called a monic morphism or a mono) is defined to be a left-cancellative morphism. It seems that this definition generalizes the definition of injections. However, even in a concrete category, a monomorphism may not be an injection. Why could this happen? I know examples of monomorphisms which are not injections but what is the reason behind the existence of such examples? Isn't monomorphism a generalised concept of injection? Similar questions can be asked about epimorphisms and surjections.

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    $\begingroup$ What more are you looking for, beyond "A generalization may not always behave exactly the same as the thing it generalized, in all respects"? For what it's worth, in the quite common case of a concrete category in which the underlying set functor is representable, (i.e., in which there is a free object on one element), all monomorphisms will be injections. In some sense, the failure of monomorphisms to be injections more generally is just the failure of the underlying set functor to always be representable. $\endgroup$ – Sridhar Ramesh Apr 9 '12 at 21:33
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    $\begingroup$ The reason behind the existence of such examples is the definition of monomorphism. Your mind is too biassed by your mathematical background. $\endgroup$ – Fernando Muro Apr 9 '12 at 21:40

There are several issues here. Let $\mathcal C$ be a category.

(1) The question only makes sense when $\mathcal C$ is concrete, as otherwise "injective" has no meaning. When $\mathcal C$ is concrete, we can choose a faithful functor $\pi : \mathcal C\to Set$, and regard a morphism $f:X\to Y$ as being injective if $\pi(f)$ is injective (i.e. a monomorphism in $Set$).

(2) However, this notion of "injective" depends on the choice of concrete structure. Indeed, if we consider one of the simplest possible categories, which has two objects $A,B$ and one nonidentity morphism $f: A\to B$, then we can concretize this category by mapping $A,B,$ to an $n$-element set and an $m$-element set, respectively, and mapping the morphism to any function between those two sets--the faithfulness condition is essentially vacuous. But now $f$ was a monomorphism (and an epimorphism, although not an isomorphism) and its concretization will not typically be any of these things, depending on how the concretization was chosen.

(3) This counterexample highlights what can go wrong. First of all the notion of monomorphism depends solely on the category, and not on a choice of concretization, so it is a better "invariant" generalization of the notion of injection from $Set$. Furthermore, it can be discussed whether or not a concrete structure exists. Finally, the phenomenon that a monomorphism can fail to be injective relative to some concrete structure is roughly attributable to a lack of "enough" morphisms and objects in the original category, so that in $Set$ one sees the map is not a monomorphism, but in the original category there are not enough objects and morphisms to find a "preimage" of any diagram exhibiting the failure of injectivity.

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    $\begingroup$ Point (2) begs the question whether every concrete category can be concretized in such a way that all monomorphisms are injective? $\endgroup$ – Emil Jeřábek supports Monica Apr 10 '12 at 11:33
  • $\begingroup$ I agree. I have posed this as a separate question. $\endgroup$ – Jack Huizenga Apr 10 '12 at 17:16
  • $\begingroup$ In regards to point (1), we do not choose a faithful functor to set, rather $\mathcal{C}$ is equipped with such a functor. $\endgroup$ – goblin Jul 3 '14 at 7:11

I would like to add some remarks (although there is already an accepted answer).

A morphism $f : X \to Y$ is a monomorphism precisely iff $f_\*(T) : \mathrm{Hom}(T,X) \to \mathrm{Hom}(T,Y)$ is injective for all $T \in C$, iff $f_\* : \mathrm{Hom}(-,X) \to \mathrm{Hom}(-,Y)$ is a monomorphism in the category of presheaves $\widehat{C}$. This leads to the following intuition: The notion of monomorphism is only well-behaved if there are enough presheaves.

If $U : C \to D$ is any functor, of course it may fail to preserve or reflect monomorphisms. However, if $U$ has a left adjoint, then it is easy to verify that $U$ preserves monomorphisms. Moreover, when $U$ is faithful, then it is easy to check that $U$ reflects monomorphisms. Thus, if $(C,U)$ is a concrete category over a category $X$ (a category $C$ equipped with a faithful functor $U : C \to X$; see Section 5 in The Joy of Cats), such that $U$ has a left adjoint, then a morphism $f$ in $C$ is a monomorphism iff $U(f)$ is monomorphism in $X$. This applies in particular to the well-behaved case that $C$ is algebraic over $X$.


Note that monomorphism is not the only way to generalise an injective map. There are stronger notions, like a section. A morphism $f:A\rightarrow B$ is a section if $\exists g:B\rightarrow A$ such that $f\circ g = 1_B$, where $1_B:B\rightarrow B$ is the identity morphism. Moreover, there also are different kinds of monomorphisms (regular monomorphisms, extremal monomorphisms), most of which are generally stronger notions than a monomorphism, but weaker than a section. The situation with epimorphisms is dual.

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    $\begingroup$ As a point of terminology, I would say that the morphism $f$ in the above has a section, in particular $g$ is one such. $\endgroup$ – Mark Grant Apr 10 '12 at 8:27
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    $\begingroup$ Mark is right. The usual name for morphisms like $f$, that have sections, is "retractions". $\endgroup$ – Andreas Blass Apr 10 '12 at 18:04
  • $\begingroup$ Section is too strong though; the function $\emptyset \rightarrow X$ is only a section if $X$ is the empty set, but it is always an injection. $\endgroup$ – goblin Jul 3 '14 at 7:13

Give a concrete category, $\pi: \mathcal{C}\to Set$, if $\pi$ has a left adjoint $L$ (it making free objects) then if $m: X\to Y$ is a monomorphisms then $\pi(m)$ is a injection i.e. a monomorphim in $Set$ (just consider the one point space $p=${$\ast$}, and two two elements of $a, b\in \pi(X)$ i.e. two maps $a, b: p\to \pi(X)$ i.e. two morphisms $a', b': L(p)\to X$, we have $\pi(m)(a)=\pi(m)(b)$ iff $m\circ a'=m\circ b'$ iff $a'=b'$ iff $a=b$). Dually, "cofree objects functor" isnt usual, anyway epimorphisms in $Set$ is a retraction, and in category setting form simple epimorphisms to retraction there is a spectra of properties: extremal , strong epimorphisms are the principals, the immersion $\iota: \mathbb{Z}\to \mathbb{Q}$ this is a epimorphisms in the cateogory of rings with unity but no a strong-Epimorphism. In $T_2$ topological spaces dense continuous maps are the epimorphisms, the extreme epimorphisms are the surjective continuos maps with co-induced topology on codomain, but no all extreme epimorphisms are retraction.


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