# Understanding the Category of open subsets of a top. space X $Op_X$

I have a problem very similar to the one posted here by Brian. I guess he also stumbled across the introduction of the category of open subsets of a topological space X (denoted by $Op_X$) in Pierre Schapira's lecture notes on Algebra and Topology (or one of his sources). There you can read:

Let $X$ be a topological space. The family of open subsets of $X$ is ordered by inclusion and we denote by $Op_X$ the associated category. Hence: $$\mathrm{Hom}_{Op_X}(U,V) =\begin{cases} \{\text{pt}\} & \text{if }U \subset V,\\ \emptyset & \text{otherwise.} \end{cases}$$

Okay, $\{\text{pt}\}$ is simply defined beforehand as a set with one single element, I get that. $U$ and $V$ are open subsets of $X$ obviously, that's okay with me, too.

What I don't get is:

1. Why "hence" - isn't that pure definition? Or is it more like "so the only possible/useful/... definition of the $\text{Hom}$-sets is: [...]"?
2. So, if $U \subset V$, we have one morphism $pt:U\to V$. I feel like everybody knows which morphism this is, but I don't see it. What does $pt$ do? Is this the identity-morphism, namely $pt(u)=u ~~\forall u\in U$ (which would include $U$ in $V$ if $U\subset V$, which would make sense somehow)? If my guess is wrong, then we must at least have some kind of unique morphism, which we can always find if $U\subset V$. But how do we get that one (or its existence)? (I somehow feel like this has got something to do with fhyve's argument about isomorphisms mentioned in the question linked above. I think I understood why we have isomorphisms in that situation, but I don't see a connection to why $Op_X$ is a category or to why/how we have that $pt$-morphism. Also another user states that $pt$ is the inclusion map. Is that correct and is the inclusion map the same as what I guessed a few lines above?)

I am new to M.SE and also relatively new to category theory, so please be kind. (: I hope my English is okay.

(I'm not sure if I should have posted this question directly to Brian's question - as I would have at a standard online board - but regarding this link from the meta, I guess this is the right way.)

I hope you guys can help me out, any effort would be much appreciated!

//Edit: As a final note, I wanted to say thank you to all contributors. I will now accept one answer (the correct/shortest answer for me would be the one about inclusion markers, but I'll accept the one that really helped me understanding what's going on with this category, hope everybody is okay with that. All of the answers and comments were really helpful and of very high quality.).

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Did you read fhyve's answer on your linked question by Brian? Because that should explain both your 1) and 2). Or is there some details missing in his explanation? – Leidem Dec 26 '13 at 20:51
Yes, of course I read fhyve's answer. I clearly stated in 2) where exactly I'm having problems with his answer. [just one second, editing...] – Piwi Dec 26 '13 at 22:00
Okay, editing took too long. :D Continued: So I guess the answer to 1) is: because the family of open subsets of X is a poset, the definition arises quite naturally from the one of a poset as a category? With 2) I'm struggeling. He's saying that pt is an arrow and no point, which is what I thought anyway. But I don't see any hint considering the question, what kind of arrow (/map) this is and more important how we can be sure to have that arrow. But probably Eric Lee's answer below is the key... – Piwi Dec 26 '13 at 22:06
1) is exactly what you write, yes. As for 2), we can view the subset relation $\subseteq$ as a subset of $X\times X$, and thus there is an arrow from $U$ to $V$ iff $(U,V)\in\subseteq$. As Eric writes, the arrow doesn't denote a function as usual, but merely states a specific connection between the objects (open sets) in the category. I suppose you could interpret the $\{pt\}$ as $\{(U,V)\}$ in this case. This point will always be there if $U\subseteq V$. – Leidem Dec 26 '13 at 22:50
Okay, then I need to understand the definition of a poset as a category. But I think I don't have a long way to go there, this whole thing slowly starts to make sense to me. (That isomorphism part of fhyve's answer on the other question is still not that clear, but I'll get that. :)) Your explanation considering the arrows corresponds with what I think I've learned from Eric's answer below and was very helpful. Thank you very much! – Piwi Dec 26 '13 at 23:34

If you have a partially order set (a poset) $X$, then you can define the category $\cal X$ whose objects are the elements in $X$ and whose morphisms are given by $$\mathrm{Hom}_{\cal X}(x,y)= \begin{cases} \{x\to y\}, & \text{if }x \le y\\ \emptyset, & \text{otherwise} \end{cases}$$ One can also define it the other way round, with a morphism $y\to x$ if $y\ge x$. It is just a matter of taste. Some people like to have morphisms going "up", other like them going "down". Of course each definition gives the opposite category of the other definition.
The family of open subsets of a topological space is a poset with $U\le V$ if $U\subseteq V$. That is why the associated category (the one induced by the order) is precisely the category descirbed in your post.
But note that in this particular case we can indeed think of the morphisms as inclusions. We could even define the category to consist of all the inclusion maps, since an inclusion $U\hookrightarrow V$ exists if and only if $U\subseteq V$.

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This sadly is not that much of a help, because if we have $x\leq y$, then $Hom(x,y)=\{x\to y\}$ which is - as far as I understand it - basically saying "the set of morphisms from x to y consists of a single morphism from x to y". That's not really much information, I wanted to understand why we only want one morphism and of what kind this morphism is. How do we know we have one? What does it do? I figured out what you said regarding the family of open sets as a poset about two hours ago in the comments section of the question, but thank you for the confirmation! :) [Continued...] – Piwi Dec 27 '13 at 0:51
The last part of your answer seems interesting. Just to make sure: if you say "define the category to consist of all the inclusion maps", you mean maps of the form $U\to V, ~x\mapsto x$? So are you saying that this arrow/morphism in the definition of $Op_X$ could indeed be seen as an inclusion map? – Piwi Dec 27 '13 at 0:59
As I said, I wanted to understand what kind of arrow this is, why we can assume that it exists and why we have exact one of these. But after collecting and putting together all the hints I got here, I think it is just a definition, namely "if we have $x\leq y$, then we set an arrow (which is not further specified in any way at all) between $x$ and $y$" (and vice versa), so the idea of that arrow as a kind of "inclusion marker" fits best. Can you agree with that? And yes, I will check how this yields a category (tomorrow) and would be happy if sb could check my thoughts then. :) – Piwi Dec 27 '13 at 17:16
Thanks. Would you or sb else please have a quick check on my thoughts considering "this definition yields a category"? (This may be too long for one comment, but I guess you can't consider it an answer to my original Q., so I post it here, hope that's okay.) Let $X$ be a poset. (1) So we got our class of objects $Ob_{\cal X}$ by taking the family of all subsets. (2) Given $x,y\in Ob_{\cal X}$, we define the morphisms as stated by Stefan above. (3) Given $x,y,z\in Ob_{\cal X}, f\in Hom(y,z), g\in Hom(x,y)$, we get by def. $x\leq y\leq z$, so $x\leq z$, so we have in $Hom(x,z)$ an unique[cont.~] – Piwi Dec 28 '13 at 15:01
[~cont.]$pt :=g\circ f$. (4) Given $x\in Ob_{\cal X}$, we have $x\leq x$, so $Hom(x,x)={pt}$, so we define this $pt:=1_x$. (A1) Associativity: Given $f\in Hom(x_1,x_2), g\in Hom(x_2,x_3), h\in Hom(x_3,x_4)$, we get $x_1\leq x_2\leq x_3\leq x_4$, so $x_1\leq x_4$, so $Hom(x_1,x_4)={pt}$ and since $(h\circ g)\circ f \in Hom(x_1,x_4)\ni h\circ (g\circ f)$, we have $(h\circ g)\circ f=pt=h\circ (g\circ f)$. (A2) units:(short form:) Since for $f\in Hom(x,y)$ we have $(f\circ 1_x), f, (1_y\circ f)\in Hom(x,y)$ and $x\leq y$, $f\circ 1_x=f=1_y\circ f$ follows by the uniqueness of $pt\in Hom(x,y)$. – Piwi Dec 28 '13 at 15:15

The answers by Stefan and Eric make good points, and I would like to add to them.
It is common to describe the notion of a category abstractly as a collection of labeled dots and arrows. This is rather like describing a group by generators and relations. You may ask "what do the morphisms do?'' but this question is essentially meaningless at this point. Just like in group theory, the group elements don't "do" anything until you pick a representation $\rho:G\rightarrow \operatorname{Aut}(V)$ of this group as a collection of linear operators on maybe some vector space.

In order to relate to your example, note first off that there are two categories in play here. Firstly, there's $Op_X$ that you defined in your question. Secondly, there's the category $Op_X'$ with the same set of objects (open sets of $X$) and morphisms given by the inclusion maps. It might be helpful to think of $Op_X$ as the abstract category and $Op_X'$ as a particular representation of $Op_X$. In this case, there is an obvious isomorphism $Op_X\cong Op_X'$, so you'd be excused for identifying them in your mind. But in many cases it is better to distinguish between the "abstract" object and its "concrete" realization, whatever this may mean.

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I've heard that "category=labeled dots and arrows"-thing as categories were introduced, yes. I thought that was just some handwaving for introduction's sake. But it turns out to be a quite good way of looking at things here, I get that now. Thanks for that group theory analogy, adds a bit to my imagination. On $Op_X$/$Op_X'$: I asked this already somewhere above, but just to make sure again - you are saying I may see this arrow in the def. of $Op_X$ as the inclusion map ($U\to V, ~x\mapsto x$) (keeping in mind that I'm essentially switching between isomorphic categories here)? – Piwi Dec 28 '13 at 9:53

I think the answer is that morphisms do not have to correspond to functions, and in this case pt does not need to correspond to any function - it's just kind of a marker which tells you about an inclusion.

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Well, that sounds kind of weak and too simple. I know that morphisms are mere "arrows", yet I thought (and until today, I have not come across a different example) they always correspond to some kind of maps. If you are right, the morphisms in that category don't really do much, that's a bit strange. But I think I will try to get along with this morphisms = inclusion-markers point of view and see how far I can get. Thank you so far! – Piwi Dec 26 '13 at 22:15
It isn't a weak notion. It is a general notion. The power of category theory is that it describes a very general vocabulary for dealing with mathematical structures, specifically by encoding that structure in terms of objects and the morphisms between them, and imposing constraints on the "pictures", in terms of commutative diagrams. – nomen Jan 14 '14 at 22:35
Yes, I'm sorry. "weak" is not the right word here, I just wanted to express that - back then - I exptected morphisms to do more than just being a kind of on/off-switch (since I was used to morphisms = maps). Because I wanted to understand the category, I wanted to understand the morphisms and in order to do that, I needed to know what they do. But by now, I understand that they don't really do that much, it's a general notion. Why they are defined the way they are makes much more sense after checking the category definition on $Op_X$ and having a closer look at (pre-)sheaves. – Piwi Jan 19 '14 at 1:28