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 $\tau = \bigcup_i \varphi_i (\tau_i)$?
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..