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Let $(X_\alpha)_{\alpha \in A}$ be a collection of topological spaces. Their disjoint union $X:=\bigsqcup\limits_{\alpha \in A} X_\alpha$ is defined as the set of all ordered pairs $(x, \alpha)$ with $\alpha \in A, x \in X_\alpha$. The disjoint union topology is defined to be the one where sets are open if and only if their intersection with each $X_\alpha$ is open in $X_\alpha$.

I have checked that this is a topology and that it has the property (called the characteristic property) that for any topological space Y: $$ (f: X \rightarrow Y \textrm{ is continuous} )\Leftrightarrow ( \left. f \right\rvert_{X_\alpha} \textrm{ is continuous for each } \alpha \in A ) $$

I'm trying to prove that the disjoint union topology is the unique topology with the characteristic property. Let $\tau_d$ be the disjoint union topology and $X_d$ be the disjoint union equipped with it, as well as $\tau_s$ be another topology with the characteristic property, and $X_s$ the disjoint union equipped with $\tau_s$.

Because $\tau_d$ has the characteristic property, to prove that the identity $Id_{ds}: X_d \rightarrow X_s$ is continuous, it is enough to see that $Id_{ds}|_{X_\alpha}$ is continuous for any $\alpha$. Similarly, because $\tau_s$ has the characteristic property, to prove that the identity $Id_{sd}: X_s \rightarrow X_d$ is continuous, it is enough to see that $Id_{sd}|_{X_\alpha}$ is continuous for any $\alpha$.

This is where I am stuck. For example, if $U \in \tau_s$, then $\left( Id_{ds}|_{X_\alpha}\right)^{-1}(U) = U \cap X_\alpha$, but I cannot conclude that this set is open in $X_\alpha$ with the topology induced by $\tau_d$ (intersections of all sets in $\tau_d$ with $X_\alpha$).

The other way around, if $U \in \tau_d$, then $\left( Id_{sd}|_{X_\alpha}\right)^{-1}(U) = U \cap X_\alpha$. $U \cap X_\alpha$ is indeed open in $X_\alpha$ by definition of $\tau_d$, but I would need it to be open in the subspace topology induced by $\tau_s$ to $X_\alpha$ (which might not coincide with the initial topology on $X_\alpha$) in order to know that $Id_{sd}|_{X_\alpha}$ is continuous.

I'm getting the feeling that maybe I'm reading wrongly what it means for $f|_{X_\alpha}$ to be continuous? But how can it be anything else?...

Thank you for your time, I appreciate any help.

rosecabbage
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What you’re overlooking is that each $X_\alpha$ already has a topology $\tau_\alpha$. If $\iota:X\to X$ is the identity map, then $\iota\upharpoonright X_\alpha$ is the identity map on $X_\alpha$ for each $\alpha\in A$, which is certainly continuous with respect to $\tau_\alpha$ irrespective of the topology on $X$. Both $X_d$ and $X_s$ have the characteristic property, so $\iota$ is continuous whether viewed as a map from $X_s$ to $X_d$ or as a map from $X_d$ to $X_s$.

Brian M. Scott
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    Thank you for your reply. I understand why $i |{X\alpha}$ is continuous with respect to $\tau_\alpha$ when the topology on $X$ has the characteristic property, but I don't under stand how that helps me, because I need to show that $i |{X\alpha}$ is continuous not with respect to $\tau_\alpha$ on $X_\alpha$, but to the subspace topology induced on $X_\alpha$ by a topology with the characteristic property for $X$ (don't I?). I'm sorry if it's something completely obvious that I'm missing. – rosecabbage Aug 17 '20 at 19:46
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    @rosecabbagedragon: No, you don’t. What you’re missing here is that in the statement of the characteristic property the subspaces are assumed to have their given topologies, the ones that I called $\tau_\alpha$, irrespective of the topology on the set $X$. – Brian M. Scott Aug 17 '20 at 19:49
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    I think I understand now. So by definition, one of the equivalent conditions in the characterstic property is that $f|{X\alpha}$ is continuous with respect to $\tau_\alpha$, and not to the subspace topology induced on $X_\alpha$. Thank you, my textbook failed to mention that. – rosecabbage Aug 17 '20 at 20:00
  • @rosecabbagedragon: You’re welcome. – Brian M. Scott Aug 17 '20 at 20:04
  • Thank you rosecabbagedragon and @Brian M. Scott for your posts. I had the same exact question. Just to make sure I understand correctly, the theorem says $f: X \to Y$ is continuous iff each $f|{X{\alpha}}: X_{\alpha} \to f(X_{\alpha})$ is continuous, where the domain $X_{\alpha}$ is equipped with whatever the original topology on $X_{\alpha}$ is? Then what about the co-domain of the restriction? Is $f(X_{\alpha})$ assumed to be equipped with the subspace topology on $X$? – Leonidas Nov 19 '22 at 21:39
  • Oops, I meant "with the subspace topology on $Y$" in the last line. – Leonidas Nov 19 '22 at 21:56
  • @BrianM.Scott: I'm still confused however, probably because of the notation. I'm reading Introduction to Topological Manifolds by John Lee, and the author initially distinguishes between the sets $X_{\alpha}$ and $X_{\alpha}^{} := \iota_{\alpha}(X_{\alpha})$ (where $\iota_{\alpha}: X_{\alpha} \to \bigsqcup_{\alpha \in A} X_{\alpha}$ is given by $\iota_{\alpha}(x) = (x,\alpha)$) before quickly discarding the notation, and just identifying $X_{\alpha}$ with $X_{\alpha}^{*}$. Being a topology newbie, I'm finding this to be rather confusing and ambiguous... – Leonidas Nov 20 '22 at 19:43
  • When you say "the subspaces are assumed to have their given topologies, the ones that I called $\tau_{\alpha}$, irrespective of the topology on the set," are you referring to the topologies on the literal $X_{\alpha}$'s, or on the $X_{\alpha}^{*}$'s? – Leonidas Nov 20 '22 at 19:47
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    @Leonidas: (Sorry to have been so slow.) Yes, the topology on $f[X_\alpha]$ is the subspace topology inherited from $Y$. As for your second question, if we’re being very careful, we replace $X_\alpha$ with $X_\alpha^$ and $\tau_\alpha$ by $\tau_\alpha^={U\times{\alpha}:U\in\tau_\alpha}$. (The whole point of the starred stuff is to ensure that we’re taking a union of pairwise disjoint sets. You’ll find that in practice people are often a bit sloppy and simply act as if the sets $X_\alpha$ were pairwise disjoint; this has the advantage of simplifying the notation, and in many ... – Brian M. Scott Nov 20 '22 at 22:13
  • ... contexts it turns out to be pretty harmless.) – Brian M. Scott Nov 20 '22 at 22:13
  • @BrianM.Scott: Thanks for your reply! (And sorry for resurrecting a 2 year old post :)). Let me see if I understand what's going on: Using the OP's notation, for the first part, we want to show that $\text{Id}|{X{\alpha}^{}}: X_{\alpha}^{} \to X_{\alpha}^{}$ is continuous, where the domain is considered as a subspace of $(X, \tau_{s})$, and the range is considered as a subspace of $(X, \tau_{d})$? And so the topology on the range is $\tau_{\alpha}^{}$ and the topology on the domain is ${X_{\alpha}^{*} \cap V: V \in \tau_s }$? – Leonidas Nov 21 '22 at 17:20
  • (To clarify, meant to say $\text{Id}{sd}|{X_{\alpha}^{*}}$, as the OP calls it.) – Leonidas Nov 21 '22 at 17:33
  • @Leonidas: Yes, everywhere that the OP has $X_\alpha$, it really should be $X_\alpha^*=X_\alpha\times{\alpha}$, but otherwise it was okay. – Brian M. Scott Nov 21 '22 at 21:59