The Characteristic Property of Disjoint Union Spaces (Understanding theorem statement)

Question

I am reading Introduction to Topological Manifolds by Lee, and I have a question about the statement of Theorem 3.41 (Characteristic Property of Disjoint Union Spaces) on page 64:

Theorem 3.41 (Characteristic Property of Disjoint Union Spaces) Suppose that $(X_{\alpha})_{\alpha \in A}$ is an indexed family of topological spaces, and $Y$ is any topological space. A map $f: \bigsqcup_{\alpha \in A} X_{\alpha} \to Y$ is continuous if and only if its restriction to each $X_{\alpha}$ is continuous. The disjoint union topology is the unique topology on $\bigsqcup_{\alpha \in A} X_{\alpha}$ with this property.

My questions:

1. Since the $X_{\alpha}$'s are not subsets of $\bigsqcup_{\alpha \in A} X_{\alpha}$, the statement the "$f: \bigsqcup_{\alpha \in A} X_{\alpha} \to Y$ is continuous if and only if its restriction to each $X_{\alpha}$ is continuous" cannot be literally true. In a preceding paragraph, the author says,

For each $\alpha \in A$, there is a canonical injection $\iota_{\alpha}: X_{\alpha} \to \bigsqcup_{\alpha \in A} X_{\alpha}$ given by $\iota_{\alpha}(x) = (x,\alpha)$, and we usually identify each set $X_{\alpha}$ with its image $X_{\alpha}^{*} = \iota_{\alpha}(X_{\alpha})$.

Based on the above line about "identifying" $X_{\alpha}$ with $X_{\alpha}^{*}$, my best guess is that the author means the following: $$ f: \bigsqcup_{\alpha \in A} X_{\alpha} \to Y \text{ is continuous iff }\; f|_{X_{\alpha}^{*}}: X_{\alpha}^{*} \to f(X_{\alpha}^{*}) \; \text{is continuous for all } \alpha \in A. \hspace{1cm} (\dagger) $$ Is this the correct interpretation?

2. If $(\dagger)$ is correct, then what are the topologies on $X_{\alpha}^{*}$ and $f(X_{\alpha}^{*})$ assumed to be? Without knowing the topologies on the domain and range of $f|_{X_{\alpha}^{*}}$, the statement "$f|_{X_{\alpha}^{*}}$ is continuous" is ambiguous to me.
   

In looking at this very similar question (where they consider the case where $f$ is an identity map), according to Brian M. Scott, "the subspaces are assumed to have their given topologies." How then is the topology on $X_{\alpha}^{*}$, call it $\tau_{\alpha}^{*}$, related to the "original" topology, call it $\tau_{\alpha}$, on $X_{\alpha}$? The discussion in that post omitted the $*$ notation, which left me a little confused. My (probably wildly wrong) guess here is that $\tau_{\alpha}^{*} = \{B \times \{\alpha\}: B \in \tau_{\alpha} \}$; in other words, $\tau_{\alpha}^{*}$ is obtained by simply "tagging" all sets in $\tau_{\alpha}$ with $\alpha$.

It would help greatly to have an answer that uses the most precise notation possible (even if it's clunky!); I find the whole "identify each $X_{\alpha}$ with its image $X_{\alpha}^{*}$" to lead to ambiguity and confusion...

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Update 11/21/22: Thanks to drhab and Paul Frost for confirming that $\tau_{\alpha}^{*} = \{B \times \{\alpha\}: B \in \tau_{\alpha} \}$. Here is a (very pedantic) proof of this fact.

Proposition. Let $(X_{\alpha},\tau_{\alpha})_{\alpha \in A}$ be an indexed collection of topological spaces, let $X := \bigsqcup_{\alpha \in A} X_{\alpha}$, and let $\tau := \left\{B \subseteq X: \iota_{\alpha}^{-1}(B) \in \tau_{\alpha} \text{ for all } \alpha \in A \right\}$ be the disjoint union topology on $X$, where $\iota_{\alpha}: X_{\alpha} \to X$ is the canonical injection $\iota_{\alpha}(x) = (x,\alpha)$. Let $\tau_{\alpha}^{*}$ denote the subspace topology on $X_{\alpha}^{*} := \iota_{\alpha}(X_{\alpha})$; that is, $\tau_{\alpha}^{*} := \{ X_{\alpha}^{*} \cap B: B \in \tau \}$. Then \begin{align*} \tau_{\alpha}^{*} = \{U \times \{\alpha\}: U \in \tau_{\alpha} \}. \end{align*}

Proof. Fix $\alpha \in A$ and let $V_0 \in \tau_{\alpha}^{*}$. Then $V_0 = X_{\alpha}^{*} \cap U_0$ for some $U_0 \in \tau$. Then \begin{align*} V_0 &= \iota_{\alpha}(X_{\alpha}) \cap \iota_{\alpha}(\iota_{\alpha}^{-1}(U_0)) \\[4pt] &= \iota_{\alpha}(X_{\alpha} \cap \iota_{\alpha}^{-1}(U_0)) && (\text{since } \iota_{\alpha} \text{ is injective}) \\[4pt] &= (X_{\alpha} \cap \iota_{\alpha}^{-1}(U_0)) \times \{\alpha\}. \end{align*} Now $X_{\alpha} \in \tau_{\alpha}$ by the fact that $\tau_{\alpha}$ is a topology on $X_{\alpha}$, and $\iota_{\alpha}^{-1}(U_0) \in \tau_{\alpha}$ by the fact that $U_0 \in \tau$. Therefore, $X_{\alpha} \cap \iota_{\alpha}^{-1}(U_0) \in \tau_{\alpha}$, and so $V_0 \in \{U \times \{\alpha\}: U \in \tau_{\alpha}\}$. This proves that $\tau_{\alpha}^{*} \subseteq \{U \times \{\alpha\}: U \in \tau_{\alpha}\}$.

Now suppose $V_0 \in \{U \times \{\alpha\}: U \in \tau_{\alpha} \}$. Then $V_0 = U_0 \times \{\alpha\}$ for some $U_0 \in \tau_{\alpha}$. Now note that \begin{align*} \iota_{\beta}^{-1}(V_0) = \iota_{\beta}^{-1}(U_0 \times \{\alpha\}) = \begin{cases} U_0 & \text{if } \beta = \alpha, \\[2pt] \emptyset & \text{if } \beta \neq \alpha. \end{cases} \end{align*} Since $U_0 \in \tau_{\beta}$ when $\beta = \alpha$, and since $\emptyset \in \tau_{\beta}$ for all $\beta$, we have $\iota_{\beta}^{-1}(V_0) \in \tau_{\beta}$ for all $\beta \in A$. Therefore, $V_0 \in \tau$. It follows that $V_0 \in \tau_{\alpha}^{*}$. [This is by the following fact: If $U$ is a subspace of $S$, and $S$ is a subspace of $X$, then a subset of $S$ that is open in $X$ is also open in $S$.] This shows that $\{U \times \{\alpha\}: U \in \tau_{\alpha} \} \subseteq \tau_{\alpha}^{*}$, which completes the proof. $\qquad \square$

Answer 1

Lee defines the disjoint union space $\bigsqcup_{\alpha \in A} X_{\alpha}$ as the disjoint union set $\bigsqcup_{\alpha \in A} X_{\alpha}$ endowed with the the disjoint union topology. It is this space which occurs in Theorem 3.41. Lee does not prove anything about it, but leaves it to the reader (Exercise 3.40, Problem 3-10).

So you have to understand what the disjoint union topology looks like. Let us recall

A subset of the disjoint union is declared to be open if and only if its intersection with each set $X_\alpha$ (considered as a subset of the disjoint union) is open in $X_\alpha$.

More precisely one should say

A subset of the disjoint union is declared to be open if and only if its intersection with each set $X_\alpha^*$ is open in $X_\alpha^*$.

BUT: Which topology $\tau_{\alpha}^{*}$ do we have on $X_{\alpha}^{*}$?

Your understanding $\tau_{\alpha}^{*} = \{B \times \{\alpha\}: B \in \tau_\alpha \}$ is correct. In fact, $\tau_{\alpha}^{*}$ is the unique topology making the bijection $\iota_\alpha : X_{\alpha} \to X_{\alpha}^{*}$ a homeomorphism. This construction replaces the family of spaces $X_\alpha$ by the family of homeomorphic spaces $X_{\alpha}^{*}$ which are pairwise disjoint.

A more formal definition which avoids the use of the $X_{\alpha}^{*}$ is this:

A subset $U$ of the disjoint union is declared to be open if and only if $\iota_\alpha^{-1}(U)$ is open in $X_\alpha$ for each $\alpha \in A$.

It is then easy to see that the open subsets of the disjoint union are precisely those of the form $U = \bigsqcup_{\alpha \in A} U_{\alpha}$ with open $U_{\alpha} \subset X_\alpha$. This shows that the subspace topology of $X_{\alpha}^{*} = X_{\alpha} \times \{\alpha\} \subset \bigsqcup_{\alpha \in A} X_{\alpha}$ is in fact the above $\tau_{\alpha}^{*}$.

Let us now come to $(\dagger)$. Your interpretation is essentially correct, but there is no reason to restrict the codomain of $f \mid_{X_\alpha^*}$ to $f(X_\alpha^*)$. Simply say that "iff $f \mid_{X_\alpha^*} : X_\alpha^* \to Y$ is continuous for all $\alpha \in A$". But if you really want to work with the images $f(X_\alpha^*) \subset Y$, then they obviously receive the subspace topology from $Y$. Again a more formal statement which avoids the use of the $X_{\alpha}^{*}$ is this:

$f: \bigsqcup_{\alpha \in A} X_{\alpha} \to Y$ is continuous iff $ f \circ \iota_\alpha: X_{\alpha} \to Y$ is continuous for all $\alpha \in A$,

Answer 2

A topology $\tau$ on $\bigsqcup_{\alpha\in A}X_{\alpha}$ provides subtopologies $\tau^*_{\alpha}$ on its subsets $X^*_{\alpha}=X_{\alpha}\times\{\alpha\}$.

This by $\tau^*_{\alpha}:=\{V\cap X_{\alpha}^*\mid V\in\tau\}$.

Also it provides topologies $\tau_{\alpha}$ on the sets $X_{\alpha}$.

This by $\tau_{\alpha}=\{U\mid U\times\{\alpha\}\in\tau^*_{\alpha}\}$.

If the sets $X_{\alpha}$ and $X_{\alpha}^*$ are equipped with these topologies then the map $X_{\alpha}\to X_{\alpha}^*$ prescribed by $x\mapsto x\times\{\alpha\}$ is a homeomorphism.

You can also go backwards starting with topologies $\tau_{\alpha}$ on $X_{\alpha}$ and then construct a topology $\tau$ on $\bigsqcup_{\alpha\in A}X_{\alpha}$ by stating that a set $\bigcup_{\alpha\in A}U\times\{\alpha\}\subseteq \bigcup_{\alpha\in A}X_{\alpha}\times\{\alpha\}$ is open iff $U\in\tau_{\alpha}$ for every $\alpha\in A$.

This way there is a one-to-one correspondence between topologies $\tau$ on disjoint union $\bigsqcup_{\alpha\in A}X_{\alpha}$ and families of topologies $(\tau_{\alpha})_{\alpha\in A}$ on the sets $X_{\alpha}$ that are involved.

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Addendum:

Let me start with a less complicated situation. The result can be looked at as a lemma/claim that catches the essence of the answer.

Let $X$ be a set and let $\left(Z,\tau_{Z}\right)$ be a topological space and let $f:Z\to X$ be a function. If we define $\tau_{X}:=\left\{ B\subseteq X\mid f^{-1}\left(B\right)\in\tau_{Z}\right\} $ then it is routine to prove that $\tau_{X}$ is a topology on $X$. Now let $Y:=f\left(Z\right)\subseteq X$ and let $g:Z\to Y$ be prescribed by $z\mapsto f\left(z\right)$ so that $g$ is automatically a surjective function. Let $Y$ as a subspace of $X$ be equipped with the subspace topology. If $V$ is an open subset of $Y$ then $V=Y\cap U$ for some $U\in\tau_{X}$ and we find easily that $g^{-1}\left(V\right)=f^{-1}\left(U\right)\in\tau_{Z}$. So apparantly $g$ is also a continuous function.

Claim: If in this situation $f$ is injective then $g$ is a homeomorphism.

Proving this boils down to proving that in that situation $g$ sends open sets of $Z$ to open sets of $Y$. Let $W\in\tau_{Z}$. Then we find on base of injectivity that $f^{-1}\left(g\left(W\right)\right)=W\in\tau_{Z}$ and are allowed to conclude that $g\left(W\right)\in\tau_{X}$. This completes the prove.

Now observe the more general situation:

Let $X$ be a set. For every $\alpha\in A$ let $\left(X_{\alpha},\tau_{\alpha}\right)$ be a topological space and let $f_{\alpha}:X_{\alpha}\to X$ be a function. If we define $\tau:=\left\{ B\subseteq X\mid f_{\alpha}^{-1}\left(B\right)\in\tau_{\alpha}\text{ for every }\alpha\in A\right\} $ then it is routine to prove that $\tau$ is a topology on $X$. It is the so-called final or coinduced topology on $X$ with respect to the maps $\left(f_{\alpha}\right)_{\alpha\in A}$. It is actually the finest topology on $X$ such that these maps are all continuous. For some $\alpha\in A$ let $Y_{\alpha}=f\left(X_{\alpha}\right)\subseteq X$ and let $g_{\alpha}:X_{\alpha}\to Y_{\alpha}$ be the map that is prescribed by $x\mapsto g_{\alpha}\left(x\right)$.

Invoking the lemma above with a small adaption we find that $g_{\alpha}$ is a homeomorphism if $f_{\alpha}$ is injective. The proof is practically the same. The only difference is that by definition of $\tau$ there are more functions involved. Not only the single $f_{\alpha}$ that we focused on. This does not complicate the proof though.

This can be applied on the situation described in your question where $X=\bigsqcup_{\alpha\in A}X_{\alpha}$ and the $f_{\alpha}$ are the inclusion functions (which are of course all injective).