7
$\begingroup$

Write $\{1\}$ for the terminal category, and let $\Omega$ denote a category with two distinct (but isomorphic objects), call them $1_\Omega$ and $0_\Omega$, and a total of four arrows; two identities, an arrow $1_\Omega \leftarrow 0_\Omega$, and another arrow $0_\Omega \leftarrow 1_\Omega$.

Write $j : \Omega \leftarrow \{1\}$ for the functor given by $j(1)=1_\Omega.$

Then intuitively, the pair $(\Omega,j)$ should be able to classify the full subcategories of any other category. Explicitly, given a category $\mathbf{C}$ and a subcategory $\mathbf{S},$ there is a functor $P_\mathbf{S} : \Omega \leftarrow \mathbf{C}$ given as follows:

  • $P_\mathbf{S}(X)=1$ iff $X \in \mathbf{S}$
  • $P_\mathbf{S}(X)=0$ iff $X \notin \mathbf{S}$

The arrows go the only place they can go.

Furthermore, if we restrict to full subcategories, then this process should be reversible.

But this makes no sense!

In particular, the category $\Omega$ is equivalent to the category $\{1\}$, and hence should have no advantages over $\{1\}.$ Yet we cannot classify full subcategories using the arrow $\{1\} \leftarrow \{1\}$.

What on earth is going on here? Why do these equivalent categories seem to behave differently?

$\endgroup$
  • $\begingroup$ what exatcly means $f:X\leftarrow Y$? If it means $f:Y\rightarrow X$ I never saw this notation, is it often used in category theory? $\endgroup$ – MphLee Feb 11 '15 at 12:18
  • 1
    $\begingroup$ @MphLee, yeah that's all it means. No, its not often used; I'm just playing around with it because it makes the composition law a bit easier. Actually somewhere I read that some older category theory books use this convention, but apparently it died out. $\endgroup$ – goblin Feb 11 '15 at 12:24
  • 1
    $\begingroup$ I'm not entirely sure where is the confusion. Equivalent categories are not interchangeable. In particular, Func(A,B) and Func(A',B') can be very different (one may be finite while the other is infinite) even when A is equivalent to A' and B is equivalent to B'. Can you perhaps further explain what you mean? $\endgroup$ – Ittay Weiss Feb 11 '15 at 12:52
  • $\begingroup$ @IttayWeiss, I guess part of me believes that if we're using categories in a reasonable way, then equivalent categories should be interchangeable. Anyway, I'll add some clarification as to where I'm confused if I can work out what to write. $\endgroup$ – goblin Feb 11 '15 at 12:54
  • $\begingroup$ @Ittay: that is not true in the following sense: if $A \cong A'$ and $B \cong B'$, then $\text{Fun}(A, B) \cong \text{Fun}(A', B')$ as categories. $\endgroup$ – Qiaochu Yuan Feb 12 '15 at 2:53
9
$\begingroup$

Isomorphic categories are interchangeable in a very strong sense. Basically, whenever you have an isomorphism $A\cong B$, you can interchange $A$ for $B$ in any categorical situation. To make this more precise, just use first order logic. Let $T$ be the first order theory of categories; A $T$-model is the same as a category (add small where needed to avoid annoyances). An isomorphism of categories is the same an a $T$-isomorphism. $T$-isomorphic structures are interchangeable within the world of $T$-models and $T$-questions (i.e., problems formulated in the language of $T$ and using the axioms of $T$). This is why isomorphic categories are interchangeable within category theory; the symmetries of category theory include all permutations of isomorphic categories.

Equivalence of categories is something of a different nature. Equivalent categories can't be interchanged within category theory. The correct motto is that working inside category $A$ or inside an equivalent category $A'$ does not (essentially) matter. That is certainly true, but the emphasis here is on the word 'inside'. From the outside (but still within category theory) two equivalent categories may certainly look very different. A way to look at it is to think of a category as the category of objects modeling something. Equivalent categories offer concrete models which may be very different to what must essentially be the same thing. More precisely (and ignoring set theoretic issues again), define a 'thing' to be an equivalence class of categories (the equivalence relation is equivalence of categories). Then every category in a thing $t$, i.e., a representative of $t$, offers concrete models of $t$. Any category equivalent to it offers other models of the same $t$. It does not matter in which representative you work since you are actually interested in $t$ (the axioms defining the concrete category you actually choose for the models are just a way to turn the vague 'thing' $t$ into actual mathematical entities). However, when you can look around and see all categories and ask all sorts of categorical questions, it is not only $t$ that you suddenly see. That is why internally equivalent categories are the same, but externally they may be very different.

$\endgroup$
  • 3
    $\begingroup$ I disagree with this answer. Equivalent categories can be interchanged within category theory; if you're doing something where you can't replace the categories involved with equivalent categories, then it's not "really" category theory. $\endgroup$ – Qiaochu Yuan Feb 12 '15 at 3:13
  • 3
    $\begingroup$ Universal properties in $Cat$ determine objects up to isomorphism, not up to equivalence. I believe asking universal questions should be considered 'categorical', yet the answer is not interchangeable with any equivalent category. E.g., "is this a terminal category" is a categorical answer answered positively for any one-object-one-morphism category, but answered negatively for any other equivalent category, of which there are many. Of course, you could say that the true questions one should ask in $Cat$ should be weak 2-questions, and then the answers are equivalency invariant. $\endgroup$ – Ittay Weiss Feb 12 '15 at 3:16
  • 3
    $\begingroup$ The "correct" notion of universal property in $\text{Cat}$ is a description of the functor categories $\text{Hom}(-, X)$ (or $\text{Hom}(X, -)$), and so it should describe categories up to equivalence, not isomorphism. $\endgroup$ – Qiaochu Yuan Feb 12 '15 at 3:19
  • 1
    $\begingroup$ yes, we agree about that. You are taking the homotopical point of view, which is essentially higher categorical. Equivalently, you view $Cat$ as a 2-category (which of course it is). My answer is in the context of 1-category theory, deliberately failing to see dimension 2. It is a valid point of view I believe, particularly perhaps in the context of the relation I make with model theory. In your answer you invoke homotopy theory, which is fine unless for some reason you really need homeomorphisms as the notion of equivalency. I don't think any one of us is wrong. $\endgroup$ – Ittay Weiss Feb 12 '15 at 3:34
  • 4
    $\begingroup$ Yes, of course at some point you might be forced to do something with an actual category rather than an equivalence class of categories (e.g. to compute a $2$-limit or a $2$-colimit as an ordinary limit or colimit), but when you do so you should always recognize that you've explicitly made a choice of such a category on top of whatever else you were already doing, in the same way that when you pick a resolution of a module to compute some derived functor you should always recognize that you've chosen that resolution. $\endgroup$ – Qiaochu Yuan Feb 12 '15 at 3:39
20
$\begingroup$

Your confusion is the following: you have two categories $1$ and $\Omega$ which are equivalent, but not isomorphic. You note that for any category $C$ there is a unique functor $C \to 1$ but that functors $C \to \Omega$ seem to correspond to full subcategories.

There is a tight analogy between categories and topological spaces (so tight it can be realized by a functor, the geometric realization of the nerve) that goes like this:

  • Categories are like spaces.
  • Functors are like maps between spaces.
  • Isomorphisms of categories are like homeomorphisms of spaces.
  • Natural transformations are like homotopies.
  • Functor categories are like spaces of maps between spaces.
  • Equivalences of categories are like homotopy equivalences of spaces.

In particular, two spaces (say the point and the interval $[0, 1]$, which are quite analogous to the terminal category and $\Omega$) can be homotopy equivalent, but not homeomorphic, and hence the set of maps from some space $X$ into those two spaces can look very different. However, it's always true that if two spaces are homotopy equivalent, then the spaces of maps from $X$ to those two spaces are also homotopy equivalent: for example, the space of maps from a space $X$ into $[0, 1]$ is always contractible.

The corresponding statement for categories is that if two categories are equivalent, then the categories of functors (and natural transformations) from any other category $C$ to those categories are equivalent. In particular, we conclude that the category of functors $C \to \Omega$ must in fact be equivalent to the terminal category.

So, what are the morphisms in this category anyway? Suppose $F, G : C \to \Omega$ are two functors. What is a natural transformation between them? Well, it's a collection of maps $\eta_c : F(c) \to G(c)$ such that... but wait. In $\Omega$ there's a unique morphism between any two objects. Hence all of the $\eta_c$ are in fact uniquely determined, and automatically satisfy the necessary compatibility condition to define a natural transformation. Moreover, every such natural transformation is invertible because its inverse is also uniquely determined!

At this point let me introduce the following useful terminology: a category is contractible if there is a unique morphism between any two objects. This condition is equivalent to being equivalent to the terminal category. What we've written down is more or less a proof that if $\Omega$ is a contractible category, then so is any functor category $[C, \Omega]$.

$\endgroup$
  • $\begingroup$ How do you make the set of maps from $X$ into $[0,1]$ a topological space in its own right? $\endgroup$ – goblin Sep 13 '15 at 0:19
  • $\begingroup$ @goblin: en.wikipedia.org/wiki/Compact-open_topology $\endgroup$ – Qiaochu Yuan Sep 13 '15 at 2:19
  • $\begingroup$ I beg your pardon for commenting so long after the OP, but I am confused as to the conclusion of this answer. It seems to me what you have shown is that $\Omega$ is contractible if and only if every functor category $\Omega^{\mathbf{C}}$ is contractible, but how does this explain why $\Omega$ can characterise full subcategories, but $1$ cannot? $\endgroup$ – Rowan Aug 25 '17 at 17:30
  • $\begingroup$ @Rowan: it doesn't actually characterize full subcategories. This is an artifact of not taking seriously the principle of working up to equivalence in category theory all the time. $\endgroup$ – Qiaochu Yuan Aug 25 '17 at 17:49
  • $\begingroup$ @QiaochuYuan: there may be a more appropriate place for my line of questioning so I apologize, but could you point to important examples where the original definition of a terminal object (that is, $\text{hom}(-, X)$ has exactly one element) is less appropriate than your "correct" definition of a terminal object (that is, $\text{hom}(-, X)$ is contractible)? $\endgroup$ – Rowan Aug 28 '17 at 14:33
4
$\begingroup$

Ah, but are you really classifying subcategories? What if you have two subcategories $S$ and $S'$. This gives us the Functors $P_S$ and $P_{ S'}$, but are these functors really different? Well, we can define a natural isomorphism $!$ that for each object in $C$ is assigned the corresponding isomorphism in $\Omega$. So $P_S$ and $P_{S'}$ are basically the same Functor, because they send everything to basically the same object. It would be like distinguishing things by marking some with on pencil and others with another, technically different by essentially the same as marking them all with the same pencil.

In general, comparing Functors for equality is something called evil, precisely because it is broken under equivalence. Evil concepts generally are no different in practice from their nonevil counterparts, isomorphism in this case. See http://ncatlab.org/nlab/show/principle+of+equivalence#in_category_theory

All essentially concepts are preserved by equivalences. Just make sure not to differentiate essentially the same things, like isomorphic Functors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.