# Abstract Nonsense

## 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.

April 2, 2010 -

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 […]

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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 […]

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