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I am trying to find an example of a monomorphism that is not an injective map; much like there exist epimorphisms that are not surjective. Is this a bad question? Is every monomorphism defined on the category of sets injective? Any help would be great.

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    https://en.wikipedia.org/wiki/Monomorphism#Examples has an example of a non injective monomorphism. – Q the Platypus Dec 29 '16 at 03:27
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    Consider the subcategory of the category of abelian groups consisting of the abelian groups $\mathbb Z$ and $\mathbb Z/2\mathbb Z$, the two identity homomorphisms and all homomorphisms from the first one to the second. In this category, all morphisms are monomorphisms (and only two are injective) – Mariano Suárez-Álvarez Dec 29 '16 at 03:27
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    just ti be sure, you want a category in which objects are sets and morphisms are functions between those sets right? otherwise it is trivial. – Asinomás Dec 29 '16 at 03:27
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    @Jorge: well, otherwise it doesn't even make sense to talk about injectivity. – Qiaochu Yuan Nov 17 '18 at 17:21
  • @QiaochuYuan excellent point – Asinomás Nov 18 '18 at 17:54

2 Answers2

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In the category of divisible (abelian) groups, the map $\pi\colon\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$ is a non-injective monomorphism.

Indeed, suppose that $f,g\colon G\to \mathbb{Q}$ are such that $\pi\circ f= \pi\circ g$. Let $x\in G$. Then $f(x)-g(x) = n\in\mathbb{Z}$. If $n\neq 0$, then let $y\in G$ be such that $y^{2n}=x$ (using multiplicative notation for $G$ since we are not assuming $G$ is abelian). Then $f(x) = f(y^{2n}) = 2nf(y)$, hence $f(y) = \frac{1}{2n}f(x)$; and similarly $g(y) = \frac{1}{2n}g(x)$. Now, $f(y)-g(y)$ must be an integer, but $$f(y)-g(y) = \frac{1}{2n}(f(x)-g(x)) = \frac{1}{2},$$ a contradiction. Therefore, $n=0$, so $f(x)=g(x)$. Thus, $f=g$ and $\pi$ is a monomorphism.

One reason why non-injective monomorphisms are hard to find in algebraic contexts is that the existence of a free object on the one element set (in concrete categories) implies that monomorphisms are one-to-one, and in most "familiar" categories of algebras (in the sense of general algebra), free objects exist.

To see the implication above, let $F_x$ be the free object on the set $\{x\}$, and let $m\colon A\to B$ be a monomorphism. If $m(a) = m(a')$, then the maps $f,g\colon F_x\to A$ induced by $x\mapsto a$ and $x\mapsto a'$, respectively, satisfy $m\circ f = m\circ g$, and hence $f=g$, thus $a=a'$, proving that $m$ is one-to-one.

Arturo Magidin
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  • Nice to see you back+ – Mikasa Sep 07 '18 at 09:13
  • Maybe I'm confusing with the last part of your answer. The free object over a singleton in the unitial ring category case is the trivial ring ${0}$, so every morphism from it to another unital ring must send $0$, which is equal to $1$, to $0_R$. Then $f,g$ are not necessarily morphism. Where is my error? – John Mars Mar 06 '21 at 18:41
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    @JohnMars: No, the zero ring is terminal; it is not free. In the category of unital rings, morphisms are required to send $1$ to $1$; so there is no morphism from the zero ring to a ring with $1\neq 0$. The free object on one element in the category of unital rings is the polynomial ring $\mathbb{Z}[x]$, since $\mathbb{Z}$ is the initial object. – Arturo Magidin Mar 06 '21 at 19:15
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    @JohnMars: For a ring $R$ to be a free ring on ${x}$, it must be the case that (i) there is a set-map $i\colon{x}\to R$; and (ii) for every ring $S$ and for every set-map $g\colon{x}\to S$, there exists a unique ring homomorphism $f\colon R\to S$ such that $f(i(x))=g(x)$. The zero ring just does not have that property, even in the category of all rings (where we don’t require morphisms to send $1$ to $1$). – Arturo Magidin Mar 06 '21 at 19:18
  • @ArturoMagidin: I realise that this is quite an old answer, but I am having trouble understanding the parenthetical "using multiplicative notation because we are not assuming $G$ is abelian". To prove that a morphism $\pi:A\to B $ is a monomorphism, I believe you have to show that for every object $G$ in the category, and morphisms $f,g: G\to A$, if $\pi\circ f = \pi\circ g$, then $f=g$. Here, the category consists of divisible abelian groups, so I thought that we can assume $G$ is a divisible abelian group, since otherwise it does not belong to the category. Where have I gone wrong? – Joe Jul 21 '23 at 17:16
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    @Joe the example works in the category of divisible groups, and also in the category of divisible abelian groups, The same proof works in both categories. – Arturo Magidin Jul 21 '23 at 17:53
  • @ArturoMagidin: Okay, thanks for clarifying. I misread the phrase "divisible (abelian) groups" – I thought the parenthetical "abelian" was there to signify that your divisible groups are abelian by definition (since this is the convention adopted by some authors). But now I understand that you meant something completely different. – Joe Jul 21 '23 at 18:24
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Consider the category of pointed, connected, locally connected and locally path-connected spaces. Any nontrivial covering map is a monomorphism in this category which is not injective on underlying sets; this is a restatement of one of the lifting properties of covering maps.

Qiaochu Yuan
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