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Similar to my previous question about quotient spaces, I need to understand the definition and construction of the disjoint union topology. As before, definitions in my textbooks and online has felt too handwavy for me to understand what exactly the disjoint union topology looks like, and what the construction of such a topology would look like. There is no need to read this Wikipedia article, but for reference it is where I am getting my definition and notation. My question is split into three sub-parts, using a toy example to try and illustrate which parts of the construction I feel I need to know more about:

Let's first construct a disjoint union between sets (coproduct in $\mathrm{Set}$):

Have $i \in (I = \{1, 2 \})$.

Let $X_1 = \{a, b, c \}$ and $X_2 = \{a, b \}$ be sets.

Then the disjoint union is defined as $X = \coprod_i X_i = X_1 \sqcup X_2$

For each $i \in I$, let $$ X_i \xrightarrow{\quad \varphi_i \quad} X$$ be the canonical injection defined by

$$X_i \ni x \overset{\varphi_i}{\mapsto} (x, i)$$

The disjoint union topology on is defined to be the finest topology on for which all the canonical injections are continuous. That is, it is the final topology on induced by the canonical injections.

In this instance, we have that our set $X = \{ (a, 1), (b, 1), (c, 1) , (a, 2), (b, 2) \} $

(a) How different can the morphism $\xrightarrow{\quad \varphi_i \quad}$ be for each $X_i$? There seems to be a strict rule for what each $\varphi_i$ can do. Is indexing $\varphi_i$ superfluous notation?

The next part I find most confusing, but I'll make an attempt:

We'll now attempt to construct the disjoint union topology on $X$. I'll attempt to induce it using a discrete topology that "works" both $X_i$'s. Let's name the discrete topology $\tau_1$ for space $(X_1, \tau_1)$ and name the discrete topology $\tau_2$ for space $(X_2, \tau_2)$. Then:

$$ \tau_1 = \{ \{X_1 \}, \{\emptyset \}, \{a \}, \{b \}, \{c \}, \{a, b \}, \{a, c\}, \{b, c \} \} $$

$$ \tau_2 = \{ \{X_2 \} , \{\emptyset \} , \{a \}, \{b \} \} $$

Let's name some topology $\tau$ for space $(X, \tau)$. To make sure $\varphi_i$ is continuous for some space $(X_i, \tau_i)$, we check the condition that:

$$ * \in \tau \Longrightarrow \varphi^{-1}_i (*) \in \tau_i $$

For the inverse of the canonical injection defined by:

$$ (x, i) \overset{\varphi^{-1}_i}{\mapsto} x \in X_i$$

Looking at this, it seems to be that if both $\tau_1$ and $\tau_2$ are discrete then the induced topology $\tau$ looks to me like the union of two discrete topologies $\varphi_1 (\tau_1) \cup \varphi_2 (\tau_2)$:

$$ \tau = \{ \emptyset, \{(a, 1)\}, \{(b, 1)\}, \{(c, 1)\}, \{(a, 1), (b, 1)\}, \{(a, 1), (c, 1)\}, \{(b, 1), (c, 1)\}, \{(a, 1), (b, 1), (c, 1)\}, \{(a, 2)\}, \{(b, 2)\}, \{(a, 2), (b, 2)\} \} $$

Which I think ensures that each $\varphi_i$ is continuous

b) True or false? Each topology $\tau_i$ MUST be the same.

I'll assume that (b) is false for now: Let's tweak things a little. Let's have $\tau_1$ now be the trivial topology:

$$ \tau_1 = \{ \{X_1 \}, \{\emptyset \} \} $$

Then the topology appears to be the union of $$ \tau = \{ \emptyset, \{(a, 1), (b, 1), (c, 1)\}, \{(a, 2)\}, \{(b, 2)\}, \{(a, 2), (b, 2)\} \} $$

c) Is the topology $\tau = \bigcup_i \varphi_i (\tau_i)$?

Nate
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    You are thinking much too granularly about this. First have the topological picture clear in your mind: topological spaces are like little blobs, and the disjoint union topology is what happens when you take one of those little blobs and put it next to another one, without them touching (that's what makes it a disjoint union). So e.g. it should be clear, without needing to work through the details, that a convergent sequence in the disjoint union topology is a sequence which eventually becomes a convergent sequence in one of its components, that sort of thing. – Qiaochu Yuan Sep 16 '23 at 19:20
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    It should also be clear, again without needing to work through the details, that the disjoint union of discrete spaces is another discrete space (a discrete space is just a bunch of little dots and you're just putting two bunches of little dots together and you get a bigger bunch of dots). If this isn't clear then you haven't understood what the point of this construction is and I would suggest thinking about that before trying to write down the nitty-gritty details. – Qiaochu Yuan Sep 16 '23 at 19:21

1 Answers1

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For (a), functions are defined as having a fixed domain and codomain; you can't use the same function with different domains. The inclusion maps allow you to refer to which part of the disjoint union correspond to whichever input to the disjoint union that you want to talk about.

I can say that the discrete space $\{4,5,6\}$ is a disjoint union of the discrete spaces $\{0,1\}$ and $\{0\}$, but there are 6 different "ways this could happen"--correspondences between the component spaces and the disjoint union. The disjoint union from your recipe (you might call it a "tagged union") would be something like $\{(0, L), (1, L), (0, R)\}$ (L for "from the left set" and R for "from the right set") with the discrete topology, and that works great as a recipe for constructing disjoint unions, but we also want to be able to identify disjoint unions "in the wild", that don't look exactly like this as sets, but are homeomorphic, like $\{4,5,6\}$.

(b) is False: you can take a disjoint union of any space with any other space, or even disjoint unions of any set of spaces.

(c) is not quite right, since $(0.5, 0.6) \cup (2.5, 2.6)$ ought to be open in the disjoint union $[0,1] \sqcup [2,3]$, but it's not open in either of the component intervals. You want to be taking unions of the actual open sets from different pieces, not just unions of the collections of open sets..

Dennis
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  • Thanks for the answer Dennis, but I haven't seen the notation you're using for ${4, 5, 6 } $ or anything involving decimals $(0.5, 0.6) \cup (2.5, 2.6)$. Could you clarify what that means in your notation? – Nate Sep 16 '23 at 19:31
  • ${4,5,6}$ is a set with three elements. (0.5, 0.6) is the open interval, i.e., the set of real numbers greater than 0.5 and less than 0.6. – Dennis Sep 16 '23 at 19:33
  • Yeah I just realized the issue. How do you even define continuity on this topology then? Let's abbreviate for my example any ordered pair $(\star, 1)$ to $\star_1$ for brevity. How can you get continuity when $\varphi^{-1}_1 {a_1, b_1, c_1, a_2, b_2}$ quite literally can't operate on $a_2, b_2$? Do we exploit the fact that it isn't a function and have it only operate on the elements that are some $\star_1$ in $X$? Is that even allowed? – Nate Sep 16 '23 at 19:47
  • If $f\colon X \to Y$ is a function and $U\subseteq Y$, remember that the pre-image $f^{-1}(U)$ is defined as ${x \in X,:,f(x)\in U}$. So $\varphi_1^{-1}({a_1, b_1, c_1, a_2, b_2})$ only cares about $a_1$, $b_1$, and $c_1$, ignoring $a_2$ and $b_2$. – Dennis Sep 16 '23 at 20:06
  • Awesome, I can see that the case for (b) then would be continuous if the topology were $\tau={\emptyset,{a_1},{b_1},{c_1},{a_2},{b_2},{a_1,b_1},{a_1,c_1},{a_1,a_2},{a_1,b_2},{b_1,c_1},{b_1,a_2},{b_1,b_2},{c_1,a_2},{c_1,b_2},{a_2,b_2},{a_1,b_1,c_1},{a_1,b_1,a_2},{a_1,b_1,b_2},{a_1,c_1,a_2},{a_1,c_1,b_2},{a_1,a_2,b_2},{b_1,c_1,a_2},{b_1,c_1,b_2},{b_1,a_2,b_2},{c_1,a_2,b_2},{a_1,b_1,c_1,a_2},{a_1,b_1,c_1,b_2},{a_1,b_1,a_2,b_2},{a_1,c_1,a_2,b_2},{b_1,c_1,a_2,b_2},X} $ Thanks man! – Nate Sep 16 '23 at 20:30