Conditions for "monic $\iff$ injective"?

Question

I'm trying to better understand the relationship between monicity and injectivity as properties of a morphism. My question has two parts.

First,

what are the necessary and sufficient conditions in a category for every monic morphism to be injective?

(I'm guessing that this is not an open question!)

I am also interested in the reverse implication (i.e. every injection is monic). In this case, the standard definition of injective (i.e. $f(x) = f(y) \Rightarrow x = y$) already implies that every injective morphism is monic, but this definition is "pointwise", which is not a very satisfying as a categorical criterion. Therefore, the second part of my question is

is there a way to define injective that does not make any reference to individual elements of sets, but yet is equivalent to the "standard" definition?

(My hope is that from such a point-free definition I will be able to come up with a point-free, purely categorical proof of "injective $\Rightarrow$ monic".)

Thanks!

P.S. This question was motivated by some remarks on p. 32 of Sets for Mathematics by Lawvere and Rosebrugh (2003), where they define what a monomorphism is. Basically their definition says that a morphism $i:S \rightarrow A$ is monomorphic iff, for every pair of morphisms $s_1, s_2:T \rightarrow S, (i \circ s_1 = i \circ s_2) \Rightarrow (s_1 = s_2)$, which they call the "cancellation property with respect to composition on [the morphism's] right". Then they go on to say that

The difference between monomorphic and injective is that, for the "mono" property, we require cancellation for all $T$ [whereas for the 'injective' property, we require cancellation only when composing with morphisms with domain $T = 1$]. This does not matter in the case of abstract sets, where cancellation with the general $T$ or with just $T > = 1$ means the same thing ... . *That "mono" implies injective is * *tautologous because a general statement always implies any of its special * cases. The converse statement is $not$ tautologous; it depends on the existence of sufficiently many elements. (emphasis added)

The sentence I emphasized above puzzles me, because I know of examples where a monic morphism is not injective. The following statement is also a bit puzzling, because the implication that they don't consider tautologous ("injective $\Rightarrow$ monic") is the one that looks entirely tautologous to me. And I am also puzzled by what exactly they meant by $T = 1$. Did they mean "$T =$ the category's terminal object"? Or "$T =$ the category's separator object?" Or "$T =$ a/the singleton object"? In $Sets$ all these descriptions apply to 1, but it is not clear to me which of them is the basis for the authors' comments. Furthermore, it means that the authors are either (1) implying that the term injective can be meaningfully applied only to morphisms in categories that have a "1" object (with 1 = terminal/separator/singleton, depending on what the authors meant), or else (2) implying that every concrete category has one such object.

Answer 1

First off, those comments (and the book in general) are actually about toposes, not general categories, and should be interpreted in that light. (This is something I've only come to realise recently, despite having read the book a couple of years ago.) Therefore one should look to categories of sheaves to make sense of the remarks.

And I am also puzzled by what exactly they meant by T=1. Did they mean "T= the category's terminal object"? Or "T= the category's separator object?" Or "T= a/the singleton object"? In Sets all these descriptions apply to 1, but it is not clear to me which of them is the basis for the authors' comments.

$1$ always means the terminal object in this book, if I remember correctly.

The notion of ‘singleton’ is very set-centric. It is possible to generalise this idea to toposes, but this is not what is meant here. Toposes in general do not have a separator object either; those that are separated by $1$ (and with $1$ not isomorphic to $0$) are called well-pointed.

Furthermore, it means that the authors are either (1) implying that the term injective can be meaningfully applied only to morphisms in categories that have a "1" object (with 1 = terminal/separator/singleton, depending on what the authors meant), or else (2) implying that every concrete category has one such object.

As already mentioned, not every category has a separator. Nor does every category have a terminal object. One can study the implied definition in any category with a terminal object, but it is not necessarily useful. For example, in categories where the terminal object is isomorphic to the initial object, e.g. $\mathbf{Grp}$ or $\mathbf{Ab}$, there will always be exactly one arrow $1 \to X$ for any object $X$. It is most profitable when the category is a topos. In the traditional case of categories of sheaves, an arrow $1 \to X$ is called a global section, so the comment is simply saying (for this case) that ‘a sheaf may not have enough global sections’, which is not surprising at all.

Indeed, consider the category $\mathbf{Set}^2$ consisting of pairs of sets and pairs of maps. The terminal object in this category is a pair of singleton sets. Consider the pair $(\emptyset, \{ * \})$. There is no arrow $(\{ * \}, \{ * \}) \to (\emptyset, \{ * \})$ because there is no map $\{ * \} \to \emptyset$. But clearly $(\emptyset, \{ * \})$ is not ‘empty’ (both in the intuitive sense and in the sense of not being an initial object). The book might call this an example of ‘not having enough global elements’. The way to rectify this is to consider ‘generalised elements’, which in the language of sheaves is akin to the notion of a local section. But notice that this category has a separating set, namely the pair of pairs $(\emptyset, \{ * \})$ and $(\{ * \}, \emptyset)$. It is tempting to take the coproduct of these two to try to make a single separating object, but we have already shown that doesn't work!

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Now, to answer your first question. Note that monicity can be defined as a limit: an arrow $f : A \to B$ is monic if and only if the pullback of $A \xrightarrow{f} B \xleftarrow{f} A$ is the identity map. I will discuss only concrete categories here and ‘injective’ is reserved for maps of sets.

Necessary conditions. The above characterisation is useful because we can now understand preserving monics in terms of preserving limits. Thus, if the underlying set functor of a category preserves finite limits (or even just pullbacks), the underlying map of a monic arrow must be injective. This happens, for example, when the underlying set functor has a left adjoint (the ‘free object functor’). This means, for categories of where is a reasonable notion of ‘free object generated by one element’, the underlying map of a monomorphism must be injective.

Sufficient conditions. Similarly, if the underlying set functor reflects limits, then every arrow which has an injective underlying map must already be monic. This is typical in categories of algebraic objects and is true in particular for $\mathbf{Grp}$, $\mathbf{Ring}$, $\mathbf{Vect}$. Note, however, that the underlying set functor for $\mathbf{Top}$ does not reflect (or create) limits, so one has to search for other ‘reasons’ why a monic arrow in this category is the same thing as an injective continuous map. In fact, when the underlying set functor is faithful (i.e. injective on arrows), if the underlying map of an arrow is injective, then the arrow must be monic. The easiest way to see this is to use the fact that the functor $\mathrm{Hom}(C, -) : \mathbf{C} \to \mathbf{Set}$ preserves monics. (Exercise!)

Answer 2

On page 144 of The joy of cats I encounter:

8.29 PROPOSITION If a construct A has a free object over a singleton set, then the monomorphisms in A are precisely those morphisms that are injective functions.

Answer 3

You cannot state that morphisms in a category are monic iff they are injective because morphisms are not functions in general, they're stuff that satisfies category axioms and need not be functions. For instance, a category in which your assertion doesn't make sense is in the category made out of a graph, with objects being vertices and morphisms being edges (google that for a complete definition of how that actually works out, I'm sure you can if you started reading a little of category theory).

There is something interesting about your statement though : http://en.wikipedia.org/wiki/Monomorphism

There is a way of saying that a monic morphism is "injective" in some way, if the morphisms are actually functions. In the link, there is a statement that mentions that a map $f : X \to Y$ is monic iff the induced map $f_*$ defined by $f_* : \mathrm{Hom}(Z,X) \to \mathrm{Hom}(Z,Y)$ with $f_*(h) = f \circ h$ is injective for all objects $Z$.

To prove this (the proof is not in the link), note that if $f$ is actually an injective function and the objects are sets (groups, spaces... whatever), if a map $h_1$ (which maps into $X$, and then into $Y$ by $f$) is the same thing than if $h_2$ maps into $X$ then $f$ into $Y$, this means that to show $f$ is a monic morphism, you wanna prove that $$ f_* \circ h_1 = f_* \circ h_2 \quad \Rightarrow \quad h_1 = h_2. $$ Hence, we see that $$ f(h_1(z)) = f(h_2(z)) \quad \forall z \in Z \quad \Rightarrow \quad h_1(x) = h_2(x) \quad \forall z \in Z \quad \Rightarrow \quad h_1 = h_2 $$ which means $f$ is a monic morphism. Conversely, a monic morphism is such that if in particular, $Z = X$, and $h_1$ is a morphism which maps everything to $x_1$, $h_2$ is a morphism that maps everything to $x_2$, you can see by monicity that $$ f_*(h_1) = f_*(h_2) \quad \Longleftrightarrow \quad f(x_1) = f(x_2) $$ implies that $h_1 = h_2$, hence $x_1 = x_2$.

I am not very experienced with category theory to be honest, but I think I enlightened you enough. Feel free to comment.

Hope that helps,