5
$\begingroup$

I am trying to understand the differens between the two more precisely. I am aware that the product topology is the product in category while the disjoint union topology is the coproduct. Unfortunately I feel that doesn't quite explain it. If we let $(\mathbf{A},\alpha)$ and $(\mathbf{B},\beta)$ be two given topologies. I understand that our product topology for $\mathbf{A}\times\mathbf{B}$ is that a set $(u,v)$ is open in it if and only if $u$ is open in $\mathbf{A}$ and $v$ is open in $\mathbf{B}$. So far so good, this is all crystal clear to me.

I will use the $\oplus$ symbol rather than the upside-down product sign, I don't like it. Let our disjoint union $\mathbf{A}\oplus\mathbf{B}$, a set for us here is then $u\oplus v$ and this is where I am starting to feel uncertain. I get there is a very evident inclusion map $\imath_A:\mathbf{A}\to\mathbf{A}\oplus\mathbf{B}$ which would be $\imath_A(a)=a\oplus\emptyset$, or more I deduce this from the context of what I am reading that this is the most likely candidate. I also gather from what I have read (I will inform I have not found it in a proper book in my studies so far, might have missed it but I have read this primarely online from definitions) that a set $u\oplus v$ is open if and only if $\imath_A^{-1}(u\oplus v)$ and $\imath_B^{-1}(u\oplus v)$ are open.

This doesn't strike me as really that different, or different at all, from the product topology at first sight. I will speculate a bit here so please tell me what I get right and what I get wrong. I know the product topology is the cartesian product hence no part of the pair $(a,b)$ can be empty, the empty set is of course always part in the topology by definition. What I gather from my reading though is that they often place emphasis on the word "or", as in for $w\in\mathbf{A}\oplus\mathbf{B}$ we have that either $w\in\mathbf{A}$ OR $w\in\mathbf{B}$. This leads me to conclude that if we have $a\oplus b\in\mathbf{A}\oplus\mathbf{B}$ then either $a=\emptyset$ or we have that $b=\emptyset$. In a more word friendly way, that any element in our disjoint union topology is a member of either of the components that makes up the union, and not a construct as a combination from both of the components, which it is in a proudct topology.

Am I understanding this correctly or am I missing something?

$\endgroup$
  • 1
    $\begingroup$ Not every open set in a product is a product of open sets. The disk $\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 < 1\}$ is open in the product topology (which is the usual topology induced by the Euclidean metric). A set is open in the product topology if it is a union of "open rectangles". $\endgroup$ – Daniel Fischer Oct 13 '15 at 13:43
  • $\begingroup$ That's what I meant, thanks for clearifying though :) THe product topology is not my issue though, it's the disjoint union one. $\endgroup$ – Zelos Malum Oct 13 '15 at 13:43
  • 4
    $\begingroup$ I think this comes from a misunderstanding of what the elements in $A\oplus B$ actually look like. Elements in $A\oplus B$ are not of the form $a\oplus b$. Rather they are of the form $(a,0)$ or $(b,1)$ (or some convention like that). You are not mixing $A$ and $B$. You are just putting them together in the same set, in some sense. This is the distinction. The disjoint union topology can be thought of as a union of the two different topologies on $A$ and $B$ whereas the product topology is more like a product of the topologies. $\endgroup$ – Cameron Williams Oct 13 '15 at 14:09
  • $\begingroup$ Hmmmm would that mean if we had $A\oplus B\oplus C$ that elements from $C$ would be $(c,2)$? $\endgroup$ – Zelos Malum Oct 13 '15 at 14:22
  • 1
    $\begingroup$ @ZelosMalum that's right. $\endgroup$ – Cameron Williams Oct 13 '15 at 16:28
7
$\begingroup$

Not quite. First a minor point that may or may not help to solve some of your confusion. If $a$ is a subset of $A$ and $b$ is a subset of $B$, then it does not make sense to write $a \oplus b \in A \oplus B$, but $a \oplus b \subseteq A \oplus B$ would make sense; $a \oplus b$ is not an element of $A \oplus B$ but a subset thereof. Similarly, writing $i_A(a) = a \oplus \emptyset$ is problematic, since $a \oplus \emptyset$ is not an element of $A \oplus B$. Moreover, it need not be the case that one of $a$ or $b$ is necessarily empty. For instance, if $A = \{1,2,3\}$ and $B = \{4,5\}$, then $\{1, 5\} \subseteq A \oplus B$ but both $i_A^{-1}(\{1,5\}) = \{1\}$ and $i_B^{-1}(\{1,5\}) = \{5\}$ are non-empty.

Note that this really has nothing to do with the topologies; here we are only thinking about what $A \oplus B$ is like as a set. Indeed, figuring out the difference between $A \oplus B$ and $A \times B$ as sets would be a good starting point: if they are different sets, then certainly they will also be different topological spaces.

Take $A = \{1, 2, 3\}$ and $B = \{4, 5\}$ as above. Then $A \oplus B = \{1, 2, 3, 4, 5\}$ consists of $5$ points (but depending on your treatment of disjoint unions, you may want to write this as $A \oplus B = \{(1,0), (2,0), (3,0), (4,1), (5,1)\}$ or something like that). In any case, the set will consist of $5$ elements.)

On the other hand, $A \times B = \{(1,4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)\}$ consists of $6$ points so the two sets are really different (in that they have different cardinalities and there can be no bijection (thus in particular no homeomorphism) between them).

However, coming to the actual topologies, I can see where your confusion might arise. In both the case of the disjoint union topology and the product topology we have very natural maps: in the first case we have the inclusions, and in the second case we have the projections. When we cook up the two topologies the way we do, it ensures that in the first case, the inclusions become continuous for free, and in the second case, the projections become continuous for free, so certainly there are analogues between the two notions. You already seem to be aware of this though, as you mention products vs. coproducts.

$\endgroup$
  • 1
    $\begingroup$ Ah yes, you are right about notation, that was my fault entirely and it was purely notational but you are correct on it. I am aware of quite a bit but they seem in some ways so similar that they just "are the same" in some sense except notatoin. While I get the gist of your example with $A\oplus B$ I can't help but there might be some typoes in there, I am not sure why but I just get a gut feeling of it, it helps a little to clear it out but not quite enough. $\endgroup$ – Zelos Malum Oct 13 '15 at 13:58
  • $\begingroup$ After furhter research yours win :) it makes sense now. $\endgroup$ – Zelos Malum Oct 13 '15 at 15:29
0
$\begingroup$

The difference between the two topologies all lies in their canonical projections.

Let $\{X_i:i\in I\}$ be a family of topological spaces. Consider $X=\bigsqcup{X_i}$ . Now consider the canonical injections for each $i\in I$ given by the map $f_i: X_i\to X$ given by $f_i(x)=(x,i)$. The disjoint union topology is the topology in which all the $f_i$'s are continuous.

Now the product topology: Consider $X= \prod{X_i}$. Now consider the canonical projections $p_{i}X \to X_{i}$. The product topology is that topology in which all the $p_i$'s are continuous.

The difference here lies in the canonical mappings which happen to define a topology for the product of $X_i$ and the disjoint union of $X_i$.

Also you will see that there are elements in the disjoint union topology which are open in one of the constituent unions. That is if $X= M\sqcup N$ is given the disjoint union topology then there are open set $P$ in $X$ which is open in $M$ and sets $Q$ in $X$ which are open in $N$. But in the product topology an open set has to open in both the sets.

$\endgroup$
  • $\begingroup$ I understand those things as I have a grasp on product vs coproduct but I just can't get a proper feel for what those are topologically here, that is my issue. $\endgroup$ – Zelos Malum Oct 13 '15 at 14:21
  • $\begingroup$ @ZelosMalum If you know what connectedness means then you can see that a space given the the disjoint union topology will always be disconnected . That is there will always exist a seperation. $\endgroup$ – user210387 Oct 13 '15 at 14:24
  • $\begingroup$ I know of the concept certainly and is qutie familiar, I presume the seperation goes along $\emptyset\oplus B$ and $A\oplus emptyset$ or similarly, I am writing msotly on gut feelign rather than looking it up strictly but something along those lines? If not feel free to enlighten me further as I'd love to learn even if it is not the main issue at hand. $\endgroup$ – Zelos Malum Oct 13 '15 at 14:27
  • $\begingroup$ So your actual question I am guessing is the motivation behind disjoint union topology? $\endgroup$ – user210387 Oct 13 '15 at 14:28
  • $\begingroup$ I understand the motivation as I have dealt with categories and am studying it along with homology and others. My issue is more that I have understanding of general topology and product topology, they make sense to me definitionally, but when I look at the definition for disjoint union topology, while I see the categorial distinction compared to product topology, I cannot see a topological distinction between the two and I seek to understand this distinction on that level so I don't experience this paradox where htey are equivalent in one sense but distinct in another, which I know is not. $\endgroup$ – Zelos Malum Oct 13 '15 at 14:31

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.