If $\{q_\alpha: X_\alpha \to Y_\alpha\}$ is a family of quotient maps, then $q:\coprod_\alpha X_\alpha \to \coprod_\alpha Y_\alpha$ is a quotient map.

If $\{q_\alpha: X_\alpha \to Y_\alpha\}$ is an indexed family of quotient maps, then the map $q:\coprod_\alpha X_\alpha \to \coprod_\alpha Y_\alpha$ whose restriction to each $X_\alpha$ is equal to $q_\alpha$ is a quotient map.

A subset of the disjoint union $\coprod_\alpha X_\alpha$ is open if and only if its restriction to $X_\alpha$ (considered as a subset of the disjoint union) is open in $X_\alpha$.

Proof:

We want to show that $V$ is open in $\coprod_\alpha Y_\alpha$ if and only if $q^{-1}(V)$ is open in $\coprod_\alpha X_\alpha$.

First $q$ is continuos since the restriction to $X_\alpha$ is a quotient map for each $\alpha$. So, if $V$ is open in $\coprod_\alpha Y_\alpha$ then $q^{-1}(V)$ is open in $\coprod_\alpha X_\alpha$.

Now, suppose $q^{-1}(V)$ is open in $\coprod_\alpha X_\alpha$. Then $q^{-1}(V) \cap X_\alpha$ is open in $X_\alpha$ for each $\alpha$.

But $q^{-1}(V) \cap X_\alpha=q_\alpha^{-1}(V)$. So, $V$ is open in $Y_\alpha$ for each $\alpha$.

So, $V$ is open in $\coprod_\alpha Y_\alpha$.

My concern is when I write $q^{-1}(V) \cap X_\alpha=q_\alpha^{-1}(V)$ because $q^{-1}(V) \cap X_\alpha \subset \coprod_\alpha X_\alpha$ whereas $q_\alpha^{-1}(V)\subset X_\alpha$ (as a standalone space, not as a subset of disjoint union).

How can I justify this?

Solution 1:

This will discuss some general theory on final topologies, which are a common way that topologies are defined or characterised. This is a dual version of my posting on initial topologies here, you could say, and a lot of it is analogous.

Definition:

Let $(X,\mathcal{T})$ be a topological space, and let $Y_i, (i \in I)$ be topological spaces and let $f_i : Y_i \to X$ be a family of functions. Then $\mathcal{T}$ is called the final topology with respect to the maps $f_i$ iff:

1. $\mathcal{T}$ makes all $f_i$ continuous.

2. if $\mathcal{T}'$ is any other topology on $X$ that makes all $f_i$ continuous, then $\mathcal{T}' \subseteq \mathcal{T}$.

Note that asking for the weakest such topology will always result (regardless of the maps or the topology on $Y_i$) in the indiscrete topology on $X$, as this topology makes all arriving maps continuous. So this would be a trivial notion.

When we have a situation where we already have the spaces $Y_i, (i \in I)$ with a topology and we also have set $X$ with somehow natural maps to $X$, then the final topology w.r.t. these maps is a quite natural candidate topology to put on $X$. And this can always be done, by the following:

Existence theorem for final topologies.

Let $X$ be a set and $f_i : (Y_i,\mathcal{T}_i) \to X$ be a collection of topological spaces and functions. Then there is a topology $\mathcal{T}_f$ on $X$ that is final w.r.t. the maps $f_i$. Moreover, this topology is unique and is given by: $$\mathcal{T}_f = \{O \subseteq X: \forall i \in I: (f_i)^{-1}[O] \in \mathcal{T}_i \}$$

Proof: As $$(f_i)^{-1}[\bigcup_j O_j] = \bigcup_j (f_i)^{-1}[O_j]$$ and similarly for intersections (for all indexed collections $O_j$ of subsets of $Y$ and all $i$), we see that $\mathcal{T}_f$ is closed under unions and finite intersections, as the topologies on $Y_i$ are. It's also clear that $\emptyset$ and $X$ are in $\mathcal{T}_f$, so $\mathcal{T}_f$ is indeed a topology on $X$.

Now, if $O$ is from $\mathcal{T}_f$, by definition all inverse images of $O$ under any $f_i$ are open in $Y_i$, so that indeed $\mathcal{T}_f$ makes all $f_i$ continuous. If $\mathcal{T}$ is another such topology, then let $O \in \mathcal{T}$. Then for all $i$, we have, as $f_i$ is continuous, that $(f_i)^{-1}[O]$ is open in $Y_i$. But this means exactly that $O \in \mathcal{T}_f$, so we see that $\mathcal{T} \subseteq \mathcal{T}_f$.

Unicity is clear: if $\mathcal{T}$ and $\mathcal{T}'$ are both final, then both make all $f_i$ continuous and applying property 2. to $\mathcal{T}$ we get $\mathcal{T}' \subseteq \mathcal{T}$, and applying it to $\mathcal{T}'$ gives us $\mathcal{T} \subseteq \mathcal{T}'$; hence $\mathcal{T}=\mathcal{T}'$.

Note: comparing to the case of initial topologies: there we could (in general, an exception occurs when we have one map) only give a subbase for the initial topology, not a complete description as we have here. Of course, here we needed to check that $\mathcal{T}_f$ is in fact a topology, while any collection of subsets is a subbase for some topology, without further effort.

Also, as $(f_i)^{-1}[X\setminus O] = Y_i \setminus (f_i)^{-1}[O]$, we see that the collection of closed sets of the final topology is given by $\{F \subset X: (f_i)^{-1}[F] \text{ is closed for all } i \in I \}$

Example: the quotient topology.

This is just the final topology w.r.t. a single map. If $f:X \to Y$ is some continuous map, then if $f$ is such that $Y$ has the final topology w.r.t. $f$, we say that $Y$ has the quotient topology, and that $f$ is a quotient map. This is applied to the situation where we have an equivalence relation $R$ on a topological space $X$, and $Y$ is the set of all classes of $X$ w.r.t. the relation $R$. The map $f$ (often denoted $q$) here is the one that sends $x$ to its class $[x]$. $Y$ is then also denoted by $X/R$, with the final topology w.r.t. $f$ (or $q$). In this case $f$ is onto (surjective). If we give $Y$ (under $f:X \to Y$) the quotient topology w.r.t. $f$, we have that $Y\setminus f[X]$ is a discrete subspace, because for all $A$ that are disjoint from $f[X]$ we have that $f^{-1}[A] = \emptyset$, which is open in $X$, so that all such $A$ are open under the final topology w.r.t. $f$. So this topology would trivialise the topology outside $f[X]$ anyway, which is one of the reasons that one often assumes, in the one map setting, that this one map is onto. Note that in the equivalence relation situation the standard map $q$ is automatically onto anyway.

Example: the sum topology.

Suppose we have a family $X_i, i \in I$ of topological spaces, and let $X$ be the union of all sets $X_i \times \{i\}$. ( The "$\times \{i\}$" is needed to ensure that this is a disjoint union of the $X_i$) Then we have natural maps $k_i : X_i \to X$, by sending $x \in X_i$ to $(x,i) \in X$. These maps are all 1-1, and map onto $X_i \times \{i\}$. The set $X$ with the final topology w.r.t. these maps $k_i (i \in I)$ (the standard embeddings) is called the topological sum of the spaces $X_i$, sometimes denoted by $\coprod_{i \in I} X_i$ as well.

It is easily checked that this topology is given by the subsets $O$ of $X$ such that for all $i$, the set $\{x \in X_i: (x,i)\in O\}$ is open in $X_i$. Hence all open sets are given by all unions $\bigcup_i k_i[O_i]$, where $O_i$ is open in $X_i$. Note that this makes all $k_i$ also open (and closed) maps and so $k_i: X_i \to X_i \times \{i\}$ is a homeomorphism when the latter space gets the subspace topology from $X$. It also follows that all $X_i \times \{i\}$ are themselves open (and closed) in $X$, and no non-trivial sum-space is connected.

A sort of converse also holds: if $X$ is not connected, then $X$ can be written as $A \cup B$ where $A$ and $B$ are both non-empty, open and disjoint. Then the topology of $X$ is the final topology w.r.t. the 2 embeddings $i_A: A \to X$ and $i_B: B \to X$, so $X$ is homeomorphic to the sumspace of $A$ and $B$ (where $A$ and $B$ have the subspace topology induced by $X$).

Example:

(A generalisation of the previous remark on disconnected spaces) Let $X$ be a space and let $A_i (i=1\ldots n)$ be finitely many closed (not disjoint, they're arbitary) subspaces of $X$ in the subspace topology, of course. Assume that $\bigcup_{i=1}^n A_i = X$. Then the topology on $X$ is equal to the final topology w.r.t. the inclusion maps $k_i: A_i \to X$.

Proof: $F$ is closed in the final topology w.r.t. the $k_i$, iff for all $i \in \{1,\ldots,n\}$, $(k_i)^{-1}[F]$ is closed in $A_i$. And this means, as $(k_i)^{-1}[F] = A_i \cap F$, that $F$ is closed in the final topology iff for all $i \in \{1,\ldots,n\}$ $F \cap A_i$ is closed. Now, if $F$ is closed in the original topology, then $A_i \cap F$ is closed in $A_i$, by the definition of the subspace topology, for all $i$. Hence $F$ is closed in the final topology as well. On the other hand, if $F$ is closed in the final topology, we know that $A_i \cap F$ is closed in $A_i$. Then $A_i \cap F$ is also closed in $X$ ($A_i \cap F$ is closed in $A_i$ iff there is a closed set $C$ in $X$ such that $A_i \cap F = A_i \cap C$, and as $A_i$ is closed in $X$, and $C$ too, the intersection $A_i \cap F$ is also closed in $X$) and $$F = F \cap (\bigcup_i A_i) = \bigcup_i (A_i \cap F)$$ is a finite union of closed sets of $X$, and thus $F$ is also closed in $X$. So these topologies indeed coincide.

Example: (the same for open subspaces)

Let $X$ be a space and let $O_i$ ($i \in I$) be any collection of open subspaces of X (in the subspace topology), and suppose that $\bigcup_i O_i = X$ (an open cover of $X$) Then the topology of $X$ is the final topology w.r.t. the inclusion maps $k_i : O_i \to X$.

Proof: as in the previous example. Now work with open sets instead of closed ones, use that an open subset of $O_i$ is open in $X$ too, and use that arbitrary unions of open sets are open.

Definition: a $T_1$ space $X$ is called a $k$-space, iff $X$ has the final topology w.r.t. all inclusion maps $i_C: C \to X$, where $C$ is a closed and compact subspace of $X$. (the $T_1$ is mostly added to ensure we always have closed and compact subspaces, namely the finite ones).

One easily shows that all first countable $T_1$ spaces and all locally compact $T_2$ spaces are examples of $k$-spaces. These spaces are important in some parts of algebraic topology and in the theory of quotient maps.

A similar definition can be made for so-called sequential spaces (an equivalent definition to the usual one). Let $X$ be a $T_2$ space, and let $\mathcal{S}$ be all sets of the form $\{x_n : n \in \mathbb{N}\} \cup \{x\}$ (all points of $X$) such that $x_n \to x$ in $X$. Then $X$ is sequential iff $X$ has the final topology w.r.t. all inclusions $i_A: A \to X$ where $A$ is from $\mathcal{S}$.

Again, as with the initial topology for maps into that space, having a final topology w.r.t. a family of maps makes it easy to see which maps starting from that space are continuous. We have the following useful:

###Universal theorem of continuity for final topologies.###

Let $X$ be a space and $f_i: Y_i \to X$ be a family of spaces and functions, such that $X$ has the final topology w.r.t. the maps $f_i$. Let $g: X \to Z$ be a function from $X$ to a space $Z$. Then

($\ast$) $g$ is continuous iff $\forall i : g \circ f_i : Y_i \to Z$ is continuous.

Proof: if $g$ is continuous, then (as property 1 of the definition of final topology gives continuity of the $f_i$) all $g \circ f_i$ are continuous as compositions of continuous maps. Suppose now that all $g \circ f_i$ are continuous, and let $O$ be any open subset of $Z$. We want to show that $g^{-1}[O]$ is open in $X$, but by the existence theorem: $g^{-1}[O]$ is open in $X$ iff for all $i \in I$ : $(f_i)^{-1}[ g^{-1}[O] ]$ is open in $Y_i$. But the latter sets are just equal to $(g \circ f_i)^{-1}[O]$, and so these sets are indeed all open, as we assumed that all $(g \circ f_i)$ to be continuous. So $g^{-1}[O]$ is open in $X$ and $g$ is continuous.

There is a converse to this as well:

###Characterisation of the final topology by the continuity theorem.###

Let $(X,\mathcal{T})$ be a space, and $f_i: Y_i \to X$ be a family of spaces and functions. Suppose that $X$ satisfies the universal continuity theorem in the following sense:

If for all spaces $Z$, and for all functions $g: X \to Z$ property ($\ast$) holds, then $X$ has the final topology w.r.t. the maps $f_i$.

Proof: the identity $1_X$ on $X$ is continuous, so applying ($\ast$) from right to left with $g = 1_X$ gives us that all $f_i$ are continuous. If $\mathcal{T}'$ is another topology on $X$ that makes all $f_i$ continuous, then consider the map $g: (X, \mathcal{T}) \to (X, \mathcal{T}')$, defined by $g(x) = x$. Then all maps $g \circ f_i$ are just the maps $f_i$ as seen between $Y_i$ and $(X, \mathcal{T}'))$ which are by assumption continuous. So by the other direction of ($\ast$) we see that $g$ is continuous, and thus (as $g(x) = x$, and thus $g^{-1}[O] = O$ for all $O$) we have that $\mathcal{T}' \subseteq \mathcal{T}$, as required for the second property of the final topology.

Application: if $q:X \to Y$, and $Y$ is a quotient space w.r.t. $q$, then a map $f: Y\to Z$ is continuous iff $f \circ q$ is continuous between $X$ and $Z$.

Application: (sum maps).

If $f_i : X_i \to Y$ is continuous, and $X$ is the (topological) sum of the spaces $X_i$, then define the sum map $f: X \to Y$ by $f((x,i))$ = $f_i(x)$. Then $f$ is continuous. This follows as $k_i \circ f = f_i$ by definition of $f$, and the universal continuity theorem.

Application: (more sum maps)

Let the maps $f_i: X_i \to Y_i$ be functions between spaces $X_i$ and $Y_i$. Let $X$ be the sum of the $X_i$ (with embeddings $k_i$) and $Y$ be the sum of the spaces $Y_i$ (with embeddings $k'_i$). Define $f: X \to Y$ by (whenever $x \in X_i$) $f((x,i)) = (f_i(x), i)$. Then $f$ is continuous iff for all $i$ in $I$: $f_i$ is continuous.

Proof: we have the relation $$\forall i \in I: f \circ k_i = k'_i \circ f_i$$

So if all $f_i$ are continuous, we see that all compositions $f \circ k_i$ are continuous, as composition of $k'_i$ and $f_i$, are continuous. And as $X$ has the final topology w.r.t. the $k_i$, we see by the universal continuity theorem that $f$ is continuous. If, on the other hand, $f$ is continuous, and $i$ is fixed, we see that f restricted to $k_i[X_i]$ maps to $k'_i[Y_i]$ by construction. As $k_i$ is a homeomorphism between $X_i$ and its image, and likewise for $Y_i$ and $k'_i$ (see above where we defined the sum topology) we have that $f_i = (k'_i)^{-1} \circ f \circ k_i$, which is thus continuous, again as a composition of continuous maps.

Application (glueing lemma for continuous maps)

Let $X$ be a space $X = \bigcup_{i=1}^n A_i$, where all $A_i$ are closed in $X$. Let $f$ be a function from $X$ to a space $Y$. Then $f$ is continuous iff for all $i \in \{1,\ldots,n\}$: $f|_{A_i}$ is continuous.

Proof: this follows from the similar example above, where we showed that $X$ has the final topology w.r.t. the inclusions $k_i: A_i \to X$. If $f$ is continuous, then all $f|_{A_i}$ are continuous, as compositions of $f$ and $k_i$, and if all $f \circ k_i = f|_{A_i}$ are continuous, the universal continuity property gives that $f$ is continuous.

Second glueing lemma for continuous maps.

Let $X$ be a space such that $X = \bigcup_i O_i$, where all $O_i$ are open subspaces of $X$. Let $f$ be a function from $X$ to a space $Y$. Then $f$ is continuous iff for all $i \in I$ we have that $f|{A_i}$ is continuous.

Proof: analogous to the previous one, based on the similar example.

Application to $k$-spaces and sequential spaces:

A map $f$ from a $k$-space $X$ to a space $Y$ is continuous iff for all compact closed subspaces $C$ of $X$, $f|_C$ is continuous.

A map $f$ from a sequential space $X$ to a space $Y$ is continuous iff for all $A in \mathcal{S}$, $f|_A$ is continuous (where $\mathcal{S}$ is defined as in the definition of a sequential space above), iff $f(x_n) \to f(x)$ in $Y$ whenever $x_n \to x$ in $X$.

As in the case of initial topologies we can state a transitive law here as well:

###Transitive law of final topologies.###

Suppose that we have a family of spaces $Y_i (i \in I)$, a space $X$, and for each $i \in I$ a collection of spaces $Z_j ( j \in J_i)$ and functions $g_{i,j}: Z_j \to Y_i$ where $j \in J_i, i \in I$, and $f_i: Y_i \to X$. Also suppose that each space $Y_i$ has the final topology w.r.t. the maps $g_{i,j}$ ($j \in J_i$). Then the following are equivalent:

A. $X$ has the final topology w.r.t. the maps $\{f_i \circ g_{i,j}: i \in I, j \in J_i\}$

B. $X$ has the final topology w.r.t. the maps $f_i$ ($i \in I$).

Proof: Suppose A holds. I will apply the characterisation of the final topology by the universal continuity theorem, so let $h: X \to T$ be an arbitrary function to some space $T$ then

1. $h$ is continuous iff $h \circ (f_i \circ g_{i,j})$ is continuous for all $i \in I$ and all $j \in J_i$. This is what A says, in essence.
2. $h$ is continuous iff $(h \circ f_i) \circ g_{i,j}$ is continuous for all $i \in I$ and all $j \in J_i$. (assocociativity of composition).
3. Applying the fact that the $Y_i$ have the final topology w.r.t. the $g_{i,j}$ we conclude $h$ is continuous iff $h \circ f_i$ is continuous for all $i \in I$.
4. the last statement is equivalent to B.

This concludes the proof.

As applications we have e.g. that compositions of quotient maps are quotient maps, that the sum of quotient maps is again a quotient map, and that the quotient $T_1$ image of a $k$-space is a $k$-space etc., etc.

We also see that direct limit spaces (for those who know them) are, as quotient spaces of sum spaces, final topologies themselves.

Note however that initial and final topologies do not mix very well:

1. the product of quotient maps need not be a quotient map.

2. A subspace of a quotient space of $X$, with restricted map, need not be the quotient of a subspace of $X$.

There are some special cases where this does hold, but I do not know of very general commutativity results, so to say, between general initial and final topologies. But I do hope to have shown that these general considerations, though not very hard, do give interesting generalisations of several well-known, seemingly isolated, facts.

Solution 2:

First $q$ is continuos since the restriction to $X_\alpha$ is a quotient map for each $\alpha$. So, if $V$ is open in $\coprod_\alpha Y_\alpha$ then $q^{-1}(V)$ is open in $\coprod_\alpha X_\alpha$.

You didn't really prove anything here. You just quoted two definitions. The proper proof goes like this:

Let $V\subseteq \coprod Y_\alpha$ be an open subset. By the defintion of the topology on the disjoint union we have $V=\coprod V_\alpha$ for some open $V_\alpha\subseteq X_\alpha$. Then

$$q^{-1}(V)=q^{-1}(\coprod V_\alpha)=\coprod q^{-1}(V_\alpha)=\coprod q_\alpha^{-1}(V_\alpha)$$

Since each $q^{-1}_\alpha(V_\alpha)$ is open then so is the right side making $q^{-1}(V)$ open.

Now for your concern:

My concern is when I write $q^{-1}(V) \cap X_\alpha=q_\alpha^{-1}(V)$ because $q^{-1}(V) \cap X_\alpha \subset \coprod_\alpha X_\alpha$ whereas $q_\alpha^{-1}(V)\subset X_\alpha$ (as a standalone space, not as a subset of disjoint union).

Let's start with formalities:

Usually if $\Lambda$ is the set of indices and $\{X_\alpha\}_{\alpha\in\Lambda}$ is a collection of topological spaces (or just sets) then the disjoint union is defined as

$$\coprod_{\alpha\in\Lambda} X_\alpha=\{(\alpha, x)\ |\ x\in X_\alpha\}\subseteq \Lambda\times\bigcup_{\alpha\in\Lambda} X_\alpha$$

Note that the existance of the usual union $\bigcup_{\alpha\in\Lambda} X_\alpha$ is guaranteed by the underlying axiomatic system (e.g. ZF). With that for a given index $i\in\Lambda$ we have a natural injection

$$L_i:X_i\to\coprod_{\alpha\in\Lambda} X_\alpha$$ $$L_i(x)=(i,x)$$

This injection naturally identifies $X_i$ with its image $\{i\}\times X_i$. And thus we usually treat $X_i$ as a subset of $\coprod_{\alpha\in\Lambda} X_\alpha$ via this identification.

With that we introduce topology on $\coprod_{\alpha\in\Lambda} X_\alpha$ via $U\subseteq \coprod_{\alpha\in\Lambda} X_\alpha$ is open if and only if $U\cap X_i$ is open in $X_i$ for each $i\in\Lambda$ or equivalently if $U=\coprod U_\alpha$ for open $U_\alpha\subseteq X_\alpha$.

(as a standalone space, not as a subset of disjoint union)

Now that you know that the standalone space can be identified with a subset of the disjoint union then that's fine. The usual set theory applies here anyway. If $A, B\subseteq X$ and $f:A\to Y$ then you can surely consider $B\cap f^{-1}(Z)$ for some $Z\subseteq Y$, why not?