Is there any kind of "completion" to obtain a category for which there are no empty hom-sets from a given category?

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    $\begingroup$ Which do you want? A category with no empty homsets, or a preorder (a category for which every homset has at most one element)? Assuming you mean the latter, then what is your question? You're aware of a functor $Cat \to PreOrd$. This functor is left adjoint to the inclusion $PreOrd \to Cat$, so if that's what you mean by "completion" then sure, it's a completion. If you mean something different by "completion", then what do you mean? $\endgroup$ – tcamps Feb 5 '15 at 18:19
  • $\begingroup$ a category with no empty homs may - in general - be very different from a preorder. A preorder may have empty homs. So , like @tcamps I suggest you clarify your question $\endgroup$ – magma Feb 6 '15 at 22:24
  • $\begingroup$ @magma. Thank you for the clarification.I do not need to have a preorder.Just having no empty hom-sets is enough. $\endgroup$ – gibarian Feb 7 '15 at 18:53
  • $\begingroup$ Do you want any other properties? Of course it's easy to come up with dumb examples (e.g. assign each category the one-object category with just the identity morphism). I am moderately confident you cannot write down a left adjoint to the inclusion of connected categories (categories with at least one morphism between any two objects) into categories. $\endgroup$ – Qiaochu Yuan Feb 7 '15 at 21:44
  • $\begingroup$ @gibarian Preorders (categories with at most one arrow in each homset) have (almost) nothing to do with having non-empty homsets (categories with at least one arrow in each homset): if a category is a preorder and has nonempty homsets, then it has exactly one arrow between each object, i.e. is an indiscrete category and is equivalent to the terminal category. Of course, the (unique) indiscrete category on the set of objects of your original category is exactly the construction that Magma discusses below. $\endgroup$ – tcamps Feb 7 '15 at 21:55

The question statement needs to be cleaned up. This question has nothing to do with preorders.

Let $Cat$ denote the category of categories and let $Cat'$ be the full subcategory of categories which have at least one arrow in each homset. The most straightforward reading of this question is

Does the inclusion $Cat' \to Cat$ have a left adjoint?

One expects the answer to be no, because the definition of $Cat'$ is rather unnatural. In fact, the answer is no, because the inclusion does not preserve limits. To show this, it suffices to specify two arrows in $Cat'$ whose equalizer (in $Cat$) does not lie in $Cat'$.

Well, let's try the simplest possible thing. Let $C$ be the category with two objects $X,Y$, and morphisms freely generated by $f: X \to Y$ and $g: Y \to X$. Let $D$ be the category with two objects $X,Y$ and morphisms freely generated by $f: X \to Y $ and $g_1,g_2: Y \to X$. For $i=1,2$, let $F_i: C\to D$ be the unique functor sending $X \mapsto X$, $Y \mapsto Y$, $f \mapsto f$, and $g \mapsto g_i$. Then the equalizer of $F_1$ and $F_2$ is the arrow category (objects $X,Y$ and a single non-identity arrow $f: X \to Y$), which does not lie in $Cat'$.

On the other hand, let $Cat_*$ denote the category of categories enriched in pointed sets. So an object is a category equipped with arrows $0_{xy} \in Hom(x,y)$ for each homset, such that $0_{xy} f = 0_{xz}$ for every $f: y \to z$ and $g0_{xy} = 0_{wy}$ for every $g: w \to x$. A morphism is a functor $F$ such that $F0_{xy} = 0_{FxFy}$. Then the obvious functor $Cat_* \to Cat$ has a left adjoint, induced by the left adjoint to the natural functor from pointed sets to sets. So if $C$ is a category, its image under the reflection $Cat \to Cat_*$ is the category $\tilde C$ where $Hom_{\tilde C}(x,y) = Hom_C(x,y) \amalg \{0_{xy}\}$ and composition defined in a way that you can hopefully guess.

We can also consider a category whose objects are similar to the objects of $Cat_*$ as described above, except without the conditions that $0_{xy}f =0_{xz}$ and $g0_{xy} = 0_{wy}$ (and morphisms as above). There is also be a left adjoint to the inclusion of this category into $Cat$: it sends $C$ to the category freely generated by $C$ along with one new morphism in each homset (by taking composites, though, we end up with lots of new morphisms in each homset).


There is a canonical construction along the lines you request:

$O: Cat\rightarrow{Set}$

the functor assigning to each small category the set of its objects, has a right adjoint

$M: Set\rightarrow{Cat}$ the functor which assigns to each set $X$ a category with objects $X$ and exactly one arrow in each hom set.

Now, think about it: in a category where you have exactly one arrow between any objects $a\rightarrow b$ and exactly one $b\rightarrow a$ and exactly one $a\rightarrow a$ and one $b\rightarrow b$ (the identities), $a\rightarrow b$ and $b\rightarrow a$ must be inverse of each other, that is $a\rightarrow b$ is an isomorphism.

Moral: all objects in the categories obtained by applying $M$ are isomorphic to each other

That's why I called this functor $M$, because it "merges" all objects in the original category into an isomorphic blob.

I am not aware of an official name for functor $M$. Please let me know if there is one?

Mac Lane's CWM 2nd edition page 90 ex. 9 describes this construction.


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