Abstract Nonsense

Crushing one theorem at a time

Disjoint Union Topology


Before we move onto the next topic in our discussion of manifolds I think it would be nice to pause and discuss a previous left out topic. Just as was the case for the product and subspace topologies the disjoint union topology (defined below).  But, before we actually define the disjoint union topology let us first describe the disjoint union.

Disjoint Union: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of non-empty sets and define the disjoint union of the collection as \displaystyle \coprod_{j\in\mathcal{J}}X_j=\bigcup_{j\in\mathcal{J}}U_j where U_j=X_j\times\{j\}.

The disjoint union can be thought of as taking a collection of sets and uniting them but keeping the lineage (where the elements came from) apparent. It can be thought of as painting each element of each X_j a different color. It should be noted though that there is no redundancy of elements in this definition. In the sense that \text{card }X_j<\infty,\text{ }j=1,\cdots,m then \displaystyle \text{card }\coprod_{j=1}^{m}X_j=\sum_{j=1}^{m}\text{card }X_j

So, now that we know what the disjoint union is let us define the disjoint union topology (D.U.T.)

Disjoint Union Topology (D.U.T.): Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of non-empty topological spaces. Define the canonical injection \varphi_j: X_j\to X where  \displaystyle X=\coprod_{j\in\mathcal{J}}X_j by x\mapsto (x,j). Then, topologize X by calling O\subseteq X open if and only if \varphi_j^{-1}(O) is open for every j\in\mathcal{J}. With this topology we call X the disjoint union topology (D.U.T.)

A universal characterization of the D.U.T. (I won’t go full into this…despite the name of the forum this isn’t category theory…yet) is the following.

Theorem: A function \displaystyle f:\coprod_{j\in\mathcal{J}}X_j\to Y is continuous if and only if f\circ\varphi_j:X_j\to Y is continuous for each j\in\mathcal{J}.

Proof: Suppose that \displaystyle f:\coprod_{j\in\mathcal{J}}\to Y is continuous and let O\subseteq Y be open. Then, f^{-1}(O) is open in \displaystyle \coprod_{j\in\mathcal{J}}X_j and so \varphi_j^{-1}(f^{-1}(O))=\left(f\circ\varphi\right)^{-1}(O) is open in X_j for each j\in\mathcal{J}. Thus, f\circ\varphi_j:X_j\to Y is continuous.

Conversely, suppose that f\circ\varphi_j:X_j\to Y is continuous for each j\in\mathcal{J} and let O\subseteq Y be open. Then, f^{-1}(O) is open in \displaystyle \coprod_{j\in\mathcal{J}}X_j since \varphi_j^{-1}(f^{-1}(O))=(f\circ\varphi_j)^{-1}(O) is open for each j\in\mathcal{J}.

The conclusion follows. \blacksquare.

The next theorem gives an idea of what the D.U.T. looks like on a class of homeomorphic spaces.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a class of topological spaces such that X_j\approx \mathcal{X}with \psi_j:X_j\to \mathcal{X} the homeomorphism. Then \displaystyle \coprod_{j\in\mathcal{J}}X_j\approx\mathcal{X}\times \mathcal{J} where \displaystyle \coprod_{j\in\mathcal{J}}X_j has the D.U.T. and \mathcal{J} the discrete.

Proof: Define \displaystyle \eta:\coprod_{j\in\mathcal{J}}X_j\to\mathcal{X}\times \mathcal{J} by (x,j)\mapsto (\psi_j(x),j).

We first prove that \eta is a bijection. So, suppose that \eta((x,j))=\eta(x',j') then (\psi_j(x),j)=(\psi_{j'}(x'),j')). It is immediately obvious that j=j'\implies \psi_j=\psi_{j'}. Thus, since \psi_j is injective \psi_{j}(x)=\psi_{j}(x')\implies x=x'. To see that it’s surjective we must merely note that if (y,j)\in\mathcal{X}\times\mathcal{J} then y\in X_j and so y=\psi_j(x) for some x\in X and so \eta\left((x,j)\right)=(\psi_j(x),j)=(y,j).

We now prove that \eta is continuous. To see this let O\subseteq \mathcal{X}\times\mathcal{J} be open. Then O=O_1\times O_2 where O_1 is open in \displaystyle \coprod_{j\in\mathcal{J}}X_j and O_2\subseteq \mathcal{J}.

We claim that \displaystyle \eta^{-1}\left(O_1\times O_2\right)=\bigcup_{k\in O_2}\psi_k^{-1}\left(O_1\right)\times\{k\}

To see this first let \displaystyle (x,m)\in\bigcup_{k\in O_2}\psi_k^{-1}(O_1)\times\{k\} then (x,m)\in\psi_m^{-1}(O)1)\times\{m\}  and so x\in\psi_m^{-1}\left(O_1\right)\text{ and }m\in O_2. Thus, \psi_m(x)\in O_1\text{ and }m\in O_2 thus (\psi_m(x),m)\in O_1\times O_2 and so \eta^{-1}\left((\psi_m(x),m)\right)=(x,m)\in\eta^{-1}\left(O_1\times O_2\right).

Conversely, let \displaystyle (x,m)\in\eta^{-1}\left(O_1\times O_2\right) then \eta\left((x,m)\right)=(\psi_m(x),m)\in O_1\times O_2 and thus \psi_m(x)\in O_1\text{ and }m\in O_2\implies x\in\psi_m^{-1}(O_1)\text{ and }m\in O_2 and so by definition \displaystyle (x,m)\in \psi_m^{-1}(O_1)\{m\}\subseteq\bigcup_{k\in O_2}\psi_k^{-1}(O_2)\times\{k\}

It easily follows that \eta^{-1}(O_1\times O_2) is a union of open sets in \displaystyle \coprod_{j\in\mathcal{J}}X_j.

A similar (and equally tedious) calculation shows that \eta is open, we omit it here. The conclusion follows. \blacksquare

It should be obvious that the canonical injections are both open and closed and thus embeddings.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a class of discrete spaces. Then, \displaystyle \coprod_{j\in\mathcal{J}}X_j under the D.U.T. is discrete.

Proof: Let \displaystyle (x,j)\in\coprod_{j\in\mathcal{J}}X_j then

\displaystyle \varphi_m^{-1}\left((x,j)\right)=\begin{cases} x & \mbox{if} \quad m=j \\ {\varnothing} & \mbox{if} \quad m\ne j\end{cases}

It follows that \varphi_m^{-1}\left((x,j)\right) is open in X_m for each m\in\mathcal{J}. Thus, (x,j) is open. The conclusion follows. \blacksquare.

Next we prove some separation invariants.

Theorem: Let \displaystyle \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of Kolomogorov spaces. Then, \displaystyle \coprod_{j\in\mathcal{J}}X_j under the D.U.T. is Kolomogorov.

Proof: Let \displaystyle (x,j),(y,k)\in\coprod_{j\in\mathcal{J}}X_j be distinct. If j\ne k then clearly X_j\times\{j\} and X_k\times\{k\} are disjoint neighborhoods of each (must stronger than what is needed). If j=k then x,y\in X_j must be distinct. But, since X_j is Kolomogorov there exists, WLOG, a neighborhood U_x which contains x but not y. Then, clearly x\in U_x\times\{j\} which is open and y\notin U_x\times\{j\}. The conclusion follows. \blacksquare.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty class of T_1 spaces. Then, \displaystyle \coprod_{j\in\mathcal{J}}X_j under the D.U.T. is T_1.

Proof: Let \displaystyle (x,j)(y,k)\in\coprod_{j\in\mathcal{J}}X_j be distinct. Just as last time if j\ne k then X_j\times\{j\} and X_k\times\{k\} are the desired neighborhoods. So, assume not. Then, x,y\in X_j are distinct and thus by assumption there exists neighborhoods U_x and U_y which contain x,y and do not contain y,x respectively. Clearly then U_x\times \{j\} and U_y\times \{j\} are the desired neighborhoods. \blacksquare.

And lastly, most interesting for our current topic of study.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of Hausdorff spaces then \displaystyle \coprod_{j\in\mathcal{J}}X_j under the D.U.T. is Hausdorff.

Proof: Let \displaystyle (x,j)(y,k)\in\coprod_{j\in\mathcal{J}}X_j be distinct. If j\ne k then X_j\times\{j\} and X_k\times\{k\} are exactly the neighborhoods of (x,j),(y,k) that we need. So assume that j=k, then x,y\in X_j are distinct. It follows from X_j‘s assumed Hausdorffness that there exists disjoint neighborhoods U_x,U_y containing x,y respectively. Clearly then U_x\times\{j\} and U_y\{j\} are the desired neighborhoods. \blacksquare.

We now discuss countability invariants of the D.U.T.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of first countable spaces then \displaystyle \coprod_{j\in\mathcal{J}}X_j is first countable under the D.U.T.

Proof: Let \displaystyle (x,k)\in\coprod_{j\in\mathcal{J}} and let \left\{B_n\right\}_{n\in\mathbb{N}} be the guaranteed countable base at x\in X_k. Define \mathfrak{B}_x=\left\{B\times\{j\}:B\in\left\{B_n\right\}_{n\in\mathbb{N}}\right\}. Clearly these are open in \displaystyle \coprod_{j\in\mathcal{J}}X_j. To see that it’s a base at x let N be a neighborhood of (x,k) in \displaystyle \coprod_{j\in\mathcal{J}}X_j. Thus, \varphi_k^{-1}(N) is a neighborhood of x in X_k. Thus, there exists some B_\ell\in\left\{B_n\right\}_{n\in\mathbb{N}} such that x\in B\subseteq \varphi_k^{-1}(N). It follows that (x,j)\subseteq B\times\{j\}\subseteq N and since B\times\{j\}\in\mathfrak{B}_x the conclusion follows. \blacksquare

Once again our next theorem will be important for the main theorem of this section.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of second countable spaces. Then, \displaystyle \coprod_{j\in\mathcal{J}}X_j is second countable if \mathcal{J} is countable.

Proof: Let \mathfrak{B}_j be the class of countable bases in each X_j and let \displaystyle \mathfrak{B}=\bigcup_{j\in\mathcal{J}}\mathfrak{D}_j where \mathfrak{D}_j=\left\{B\times\{j\}:B\in\mathfrak{B}_j\right\}. Clearly then \mathfrak{D} being the countable union of countable sets is countable. Also, each element of \mathfrak{D} is clearly open. Thus, it remains to show that \mathfrak{B} is a base. To do this let \displaystyle (x,k)\in\coprod_{j\in\mathcal{J}}X_j be arbitrary and N any neighborhood of it. Clearly then \varphi_k^{-1}(N) is a neighborhood of x in X_k. Thus, we may find some B\in\mathfrak{B}_k such that x\in B\subseteq \varphi_k^{-1}(N). It follows that (x,k)\subseteq B\times\{k\}\subseteq N. Since B\times\{k\}\in\mathfrak{B} the conclusion follows. \blacksquare

We are now at the final part of our list theorems and it should be apparent now what the piece de resistance shall be.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty class of locally Euclidean spaces of dimension n. Then, \displaystyle \coprod_{j\in\mathcal{J}}X_j is locally Euclidean of dimension n.

Proof: Let \displaystyle (x,k)\in\coprod_{j\in\mathcal{J}}X_j be arbitrary. By assumption of X_k being locally Euclidean there exists some chart (U,\psi) at x. Clearly then U\times\{k\} is a neighborhood of (x,k). Define a mapping \psi^*:U\times{k}\to\psi(U) by (y,k)\mapsto y.

Clearly this is a bijection since

\displaystyle \psi^*((y,k))=\psi^*((y',k))\implies \psi(y)=\psi(y')\implies y=y'\implies (y,k)=(y',k)

It is surjective since \psi(y)\in\psi(U) implies that y\in U which implies that (y,k)\in U\times\{k\} which then of course implies that \psi^*((y,k))=\psi(y).

Now, it is clearly continuous since \varphi(O)\subseteq \psi(U) is open implies that O is open in U and thus since U is open it is open in X_k. Thus, O\times\{k\} is open in \displaystyle \coprod_{j\in\mathcal{J}}X_j and thus lastly O\times\{k\}\cap U\times\{k\}=O\times\{k\} is open in U\times\{k\}. Similarly, O\times\{k\}\subseteq U\times\{k\} is open implies that \varphi_k^{-1}\left(O\times\{k\}\right)=O is open in X_k and thus O\cap U=O is open in U and thus finally \psi(O)=\psi^*\left(O\times\{k\}\right) is open in \psi(U).

Remembering that \psi(U) is open in \mathbb{R}^n we may conclude that (U\times\{k\},\psi^*) is a chart at (x,k). The conclusion follows by noting the arbitrariness of (x,k). \blacksquare.

We are now prepared to state our main theorem.

Theorem: Let \left\{\mathfrak{M}_n\right\}_{n\in\mathbb{N}} be countable collection of n-manifolds. Then, \displaystyle \mathfrak{M}=\coprod_{n=1}^{\infty}\mathfrak{M}_n is an n-manifold.

Proof: Combining our last couple theorem the above is justs a result since the countable union of a second countable Hausdorff locally Euclidean space of dimension n is second countable Hausdorff and locally Euclidean of dimension n. \blacksquare.

Now that we have proved the main result for this post we prove some ancillary theorems regarding the D.U.T.

Theorem: Let X_1,\cdots,X_n be finitely many topological spaces, then \displaystyle X_1\amalg\cdots\amalg X_n is compact if and only if X_j is for each j=1,\cdots,n.

Proof: First suppose that X_j,\text{ }j=1,\cdots,n is compact and let \Omega be an open cover of X_1\amalg\cdots\amalg X_n clearly then \left\{\varphi_k^{-1}(\omega)\omega\in\Omega\right\} is an open cover for X_k and so by assumption it has a finite subcover \left\{\varphi_k^{-1}(\omega_1),\cdots,\varphi_k^{-1}(\omega_m)\right\}. It clearly follows that if \Lambda_k=\left\{\omega_1,\cdots,\omega_,\right\} then \Lambda_k covers X_k\times\{k\} and so \displaystyle \Lambda_1\cup\cdots\cup\Lambda_n is a finite subcover of \Omega.

Conversely, remember that the canonical injection \varphi_k:X_k\hookrightarrow X_1\amalg\cdots\amalg X_n is an embedding and thus we must only show that X_k\times\{k\} is compact. To see this note that \displaystyle \left(X_k\times\{k\}\right)' is the union of all the X_\ell\times\{\ell\},\text{ }\ell\ne k. It is easy to see that this is open since

\displaystyle \varphi_\ell^{-1}(\left(X_k\times\{k\}\right)')=\begin{cases} X_\ell & \mbox{if} \quad \ell\ne k \\ \varnothing & \mbox{if} \quad \ell=k\end{cases}

and thus X_k\times\{k\} is closed. Thus, X_k\times\{k\} $ is a closed subspace of a compact space and thus compact. It follows from previous comment that X_k is compact.

The conclusion follows. \blacksquare

It should come as a surprise to no one that disjoint unions react very badly with connectedness.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a collection of (at least two) non-empty topological spaces. Then, \displaystyle \coprod_{j\in\mathcal{J}}X_j is disconnected.

Proof: Clearly as was stated twenty times before   \displaystyle \varnothing\subsetneq X_k\times\{k\} \subsetneq\coprod_{j\in\mathcal{J}}X_j and it’s open since

\varphi_\ell^{-1}(X_k\times\{k\})=\begin{cases} X_k & \mbox{if}\quad \ell=k \\ \varnothing & \mbox{if} \quad \ell\ne k\end{cases}

But it is also closed since

\displaystyle \varphi_\ell^{-1}\left(\left(X_k\times\{k\}\right)'\right)=\begin{cases} X_\ell & \mbox{if} \quad \ell\ne k\\ \varnothing & \mbox{if} \quad \ell =k\end{cases}

The conclusion readily follows. \blacksquare

This concludes our “short” introduction to the disjoint union topology.

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April 2, 2010 - Posted by | General Topology, Manifold Theory, Topology | , ,

2 Comments »

  1. […] spaces. Then, the coproduct of and is the set (the formal disjoint union of and ) with the disjoint union topology and the canonical injections, is the coproduct of and in […]

    Pingback by Categorical Coproducts « Abstract Nonsense | February 7, 2012 | Reply

  2. […] second way we can construct new manifolds out of old ones is via the disjoint union of topological spaces. Indeed, suppose that we have two -dimensional topological manifolds and with atlases and […]

    Pingback by Topological Manifolds (Pt. II) « Abstract Nonsense | August 30, 2012 | Reply


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