We will now change direction, and spend some time building the required theory of topological spaces, which will serve as a foundation for the rigorous definition of homology groups.
Topological spaces are a generalisation of metric spaces. Knowledge of metric spaces is mostly assumed as a prerequisite, but will be reviewed briefly here. References marked [MS] refer to Dr Roxanas’s notes for MAS331 (Metric Spaces), but equivalent results can be found in many other sources.
Let $X$ be a set. A metric on $X$ is a function $d:X\times X\to \mathbb{R}$ with properties as follows:
For all $x,y\in X$ we have $d(x,y)\ge 0$, and $d(x,y)=0$ iff $x=y$.
For all $x,y\in X$ we have $d(x,y)=d(y,x)$.
For all $x,y,z\in X$ we have $d(y,z)\le d(x,y)+d(y,z)$ (the Triangle Inequality).
A metric space is a set equipped with a metric.
We can define three different metrics on ${\mathbb{R}}^{n}$, as follows:
${d}_{1}(x,y)$ | $={\displaystyle \sum _{i=1}^{n}}|{x}_{i}-{y}_{i}|$ | ||
${d}_{2}(x,y)$ | $=\sqrt{{\displaystyle \sum _{i=1}^{n}}{({x}_{i}-{y}_{i})}^{2}}$ | ||
${d}_{\mathrm{\infty}}(x,y)$ | $=\mathrm{max}(|{x}_{1}-{y}_{1}|,\mathrm{\dots},|{x}_{n}-{y}_{n}|).$ |
Later we will explain a sense in which these are essentially the same.
Consider the set ${M}_{n}(\mathbb{R})$ of real $n\times n$ matrices (or any of the various subsets of ${M}_{n}(\mathbb{R})$ discussed in Example 1.4). We can define a metric as follows:
$\text{trace}(A)$ | $={\displaystyle \sum _{i=1}^{n}}{A}_{ii}$ | ||
$\parallel A\parallel $ | $=\sqrt{\text{trace}({A}^{T}A)}$ | ||
$d(A,B)$ | $=\parallel A-B\parallel .$ |
This is not really a new example. A little matrix algebra shows that $\text{trace}({A}^{T}A)={\sum}_{i,j=1}^{n}{A}_{ij}^{2}$, so if we identify ${M}_{n}(\mathbb{R})$ with ${\mathbb{R}}^{{n}^{2}}$ in the obvious way, then this new metric is just the same as the standard metric ${d}_{2}$ in the previous example. However, this new formula for the metric makes it easier to combine with other constructions in matrix theory, such as the definition ${O}_{n}=\{A|{A}^{T}A=I\}$ of the orthogonal group.
We will not need any examples that are much more exotic than these.
Let $X$ and $Y$ be metric spaces, with metrics ${d}_{X}$ and ${d}_{Y}$. Let $f$ be a function from $X$ to $Y$. We say that $f$ is continuous if it has the following property:
For all $x\in X$ and $\u03f5>0$, there exists $\delta >0$ such that for all ${x}^{\prime}\in X$ with $$, we have $$.
Suppose that $f$ satisfies ${d}_{Y}\mathit{}\mathrm{(}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{,}f\mathit{}\mathrm{(}{x}^{\mathrm{\prime}}\mathrm{)}\mathrm{)}\mathrm{\le}{d}_{X}\mathit{}\mathrm{(}x\mathrm{,}{x}^{\mathrm{\prime}}\mathrm{)}$ for all $x\mathrm{,}{x}^{\mathrm{\prime}}\mathrm{\in}X$. Then $f$ is continuous.
Suppose we are given $x\in X$ and $\u03f5>0$. We need to provide a number $\delta >0$ and check that it has a certain property. We just take $\delta =\u03f5$. Suppose that we have ${x}^{\prime}\in X$ with $$. By our assumptions on $f$, $x$ and ${x}^{\prime}$, we then have $$. This is the required property of $\delta $. ∎
Let $X$ be a metric space, let $a$ be a point of $X$, and let $r$ be a positive real number. We put
$$ |
and call this the open ball of radius $r$ centred at $a$.
Let $X$ be a metric space, and let $U$ be a subset of $X$. We say that $U$ is open if for every point $a\in U$, there exists $r>0$ such that $OB(a,r)\subseteq U$.
We also say that a subset $F\subseteq X$ is closed if the complement $X\setminus F$ is open.
Our definition of closed sets is not the same as in [MS, page 40], but it is equivalent to that definition, as proved in [MS, page 43].
In any metric space $X$:
The empty set and the whole set $X$ are open.
The union of any collection of open sets is open.
The intersection of any finite collection of open sets is open.
One of the main reasons for introducing metric spaces is to formalise the notion of a continuous map. The following result shows that we can do that using only the system of open sets, we do not actually need the metric.
Let $f:X\to Y$ be any map of sets, and let $B$ be a subset of $Y$. We put
$${f}^{-1}(B)=\{x\in X|f(x)\in B\},$$ |
and we call this the preimage of $B$ under $f$.
Let $f\mathrm{:}X\mathrm{\to}Y$ be any map of sets.
For any family of subsets ${B}_{i}\subseteq Y$, we have ${f}^{-1}({\bigcap}_{i}{B}_{i})={\bigcap}_{i}{f}^{-1}({B}_{i})$ and ${f}^{-1}({\bigcup}_{i}{B}_{i})={\bigcup}_{i}{f}^{-1}({B}_{i})$.
For any subset $B\subseteq Y$ we have ${f}^{-1}(Y\setminus B)=X\setminus {f}^{-1}(B)$.
If $g:Y\to Z$ is another map of sets, and $C\subseteq Z$, then we have ${(g\circ f)}^{-1}(C)={f}^{-1}({g}^{-1}(C))$.
For the first claim, we have $x\in {f}^{-1}({\bigcap}_{i}{B}_{i})$ iff $f(x)\in {\bigcap}_{i}{B}_{i}$ iff ($f(x)\in {B}_{i}$ for all $i$) iff ($x\in {f}^{-1}({B}_{i})$ for all $i$) iff $x\in {\bigcap}_{i}{f}^{-1}({B}_{i})$. All the other claims can be proved in a similar way. ∎
Let $f\mathrm{:}X\mathrm{\to}Y$ be a function between metric spaces. Then the following are equivalent:
$f$ is continuous.
For every open set $V\subseteq Y$, the preimage ${f}^{-1}(V)\subseteq X$ is an open subset of $X$.
For every closed set $G\subseteq Y$, the preimage ${f}^{-1}(G)\subseteq X$ is a closed subset of $X$.
We can now introduce the theory of topological spaces, which was mentioned briefly at the end of [MS, Section 3].
Let $X$ be a set. A topology on $X$ is a collection $\tau $ of subsets of $X$ (which are called open sets) with the following properties:
The empty set and the whole set $X$ are open.
The union of any collection of open sets is open.
The intersection of any finite collection of open sets is open.
For any set $X$, we can introduce a topology by declaring that every subset is open. This is called the discrete topology. Although this is not very interesting, it does occur naturally: it is the most natural topology on $\mathbb{Z}$ for example. At the other extreme, we can introduce a different topology by declaring that only the sets $\mathrm{\varnothing}$ and $X$ are open. This is called the indiscrete topology. Unlike the discrete topology, this is rarely relevant.
Suppose we have two different metrics on $X$, say ${d}_{\mathrm{1}}$ and ${d}_{\mathrm{2}}$. Suppose we also have positive constants ${c}_{\mathrm{1}}$ and ${c}_{\mathrm{2}}$ such that ${d}_{\mathrm{1}}\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathrm{\le}{c}_{\mathrm{1}}\mathit{}{d}_{\mathrm{2}}\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}$ and ${d}_{\mathrm{2}}\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathrm{\le}{c}_{\mathrm{2}}\mathit{}{d}_{\mathrm{1}}\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}$. Then a set $U\mathrm{\subseteq}X$ is open with respect to ${d}_{\mathrm{1}}$ iff it is open with respect to ${d}_{\mathrm{2}}$. Thus, the metrics ${d}_{\mathrm{1}}$ and ${d}_{\mathrm{2}}$ give the same topology.
Suppose that $U$ is open with respect to ${d}_{1}$. Consider a point $a\in U$. By assumption, there exists $r>0$ such that ${B}_{1}(a,r)\subseteq U$, where ${B}_{1}(a,r)$ is the open ball defined using ${d}_{1}$, or in other words
$$ |
We claim that ${B}_{2}(a,r/{c}_{1})$ is also contained in $U$. Indeed, if $x\in {B}_{2}(a,r/{c}_{1})$, then $$, so $$, so $x\in {B}_{1}(a,r)$. However, we have ${B}_{1}(a,r)\subseteq U$ by assumption, so $x\in U$. This proves that ${B}_{2}(a,r/{c}_{1})\subseteq U$ as claimed. As we can do this for any $a\in U$, we see that $U$ is open with respect to ${d}_{2}$. By a symmetrical argument, if $U$ is any set that is open with respect to ${d}_{2}$, then it is also open with respect to ${d}_{1}$. Thus ${d}_{1}$ and ${d}_{2}$ have the same open sets, and thus the same topology. ∎
The metrics ${d}_{\mathrm{1}}$, ${d}_{\mathrm{2}}$ and ${d}_{\mathrm{\infty}}$ all give the same topology on ${\mathbb{R}}^{n}$.
In the light of the lemma, it will be enough to prove the inequalities
$${d}_{\mathrm{\infty}}(x,y)\le {d}_{2}(x,y)\le {d}_{1}(x,y)\le n{d}_{\mathrm{\infty}}(x,y).$$ |
Put ${z}_{i}=|{x}_{i}-{y}_{i}|\ge 0$, then choose $p$ such that ${z}_{p}$ is the largest of all the terms ${z}_{i}$. We then have
$${d}_{\mathrm{\infty}}(x,y)=\mathrm{max}({z}_{1},\mathrm{\dots},{z}_{n})={z}_{p},$$ |
so the required inequalities are
$${z}_{p}\le \sqrt{\sum _{i}{z}_{i}^{2}}\le \sum _{i}{z}_{i}\le n{z}_{p}.$$ |
For the third inequality, the sum contains $n$ terms each of which is less than or equal to ${z}_{p}$, so the total is at most $n{z}_{p}$, as required. The other two inequalities are equivalent to
$${z}_{p}^{2}\le \sum _{i}{z}_{i}^{2}\le {(\sum _{i}{z}_{i})}^{2}=\sum _{i,j}{z}_{i}{z}_{j}.$$ |
The first sum consists of ${z}_{p}^{2}$ plus some other nonnegative terms. In the second sum, the terms with $i=j$ give the same as the first sum, and there are again some additional nonnegative terms. The claim is clear from this. ∎
For $p\in \{1,2,\mathrm{\infty}\}$ put
$${B}_{p}^{n}=\{x\in {\mathbb{R}}^{n}|{d}_{p}(0,x)\le 1\}.$$ |
The inequalities in the above proof are equivalent to the claim that $\frac{1}{n}.{B}_{\mathrm{\infty}}^{n}\subseteq {B}_{1}^{n}\subseteq {B}_{2}^{n}\subseteq {B}_{\mathrm{\infty}}^{n}$. This can be illustrated in the case $n=2$ as follows:
Let $X$ and $Y$ be topological spaces, and let $f$ be a function from $X$ to $Y$. We say that $f$ is continuous if for every open set $V\subseteq Y$, the preimage ${f}^{-1}(V)$ is open in $X$.
Proposition 3.12 shows that when $X$ and $Y$ are metric spaces with the metric topology, this agrees with our earlier definition of continuity.
We can define a function $f:{\mathbb{R}}^{4}\to {\mathbb{R}}^{4}$ by
$$f(u,v,w,x)=({e}^{u}\mathrm{cos}(x)+w\mathrm{sin}(x),-{e}^{u}\mathrm{sin}(x)+w\mathrm{cos}(x),{e}^{v}\mathrm{sin}(x),{e}^{v}\mathrm{cos}(x)).$$ |
The functions ${e}^{x}$, $\mathrm{sin}(x)$ and $\mathrm{cos}(x)$ are known to be continuous, by basic real analysis. It is also standard that sums and products of continuous functions are continuous, and it follows that each of the four components of $f$ is continuous. From this it follows easily that $f$ itself is continuous. In future, we will not bother to discuss this kind of argument in detail.
As a very basic example, consider a constant function $f:X\to Y$. This means that there is a single point $b\in Y$ such that $f(x)=b$ for all $x$. Consider an open set $U\subseteq Y$. If $b\in U$ then ${f}^{-1}(U)=X$, and if $b\notin U$ then ${f}^{-1}(U)=\mathrm{\varnothing}$. As both $X$ and $\mathrm{\varnothing}$ are open in $X$, we see that ${f}^{-1}(U)$ is always open. This proves that $f$ is continuous.
Let $f$ be as above. Then $f$ is continuous iff for every closed subset $G\mathrm{\subseteq}Y$, the preimage ${f}^{\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}G\mathrm{)}$ is closed in $X$.
Suppose that $f$ is continuous. Let $G\subseteq Y$ be closed; we must show that ${f}^{-1}(G)$ is closed in $X$. As $G$ is closed in $Y$, we know that the complement $Y\setminus G$ must be open in $Y$, so ${f}^{-1}(Y\setminus G)$ is open in $X$ (by the definition of continuity). However, ${f}^{-1}(Y\setminus G)$ is the same as $X\setminus {f}^{-1}(G)$. As this is open, we see that ${f}^{-1}(G)$ must be closed, as required.
The converse can be proved in essentially the same way. ∎
Suppose that $X$, $Y$ and $Z$ are topological spaces, and we have continuous maps $f\mathrm{:}X\mathrm{\to}Y$ and $g\mathrm{:}Y\mathrm{\to}Z$. Then the composite map $g\mathrm{\circ}f\mathrm{:}X\mathrm{\to}Z$ is also continuous. Moreover, the identity map $\mathrm{id}\mathrm{:}X\mathrm{\to}X$ is also continuous.
Consider an open set $W\subseteq Z$; we must show that the preimage ${(g\circ f)}^{-1}(W)$ is open in $X$. As $g$ is continuous, we see that ${g}^{-1}(W)$ is an open subset of $Y$. As $f$ is continuous, it follows in turn that ${f}^{-1}({g}^{-1}(W))$ is open in $X$. However, we have $x\in {f}^{-1}({g}^{-1}(W))$ iff $f(x)\in {g}^{-1}(W)$ iff $g(f(x))\in W$ iff $x\in {(g\circ f)}^{-1}(W)$, so ${f}^{-1}({g}^{-1}(W))$ is the same as ${(g\circ f)}^{-1}(W)$. We have therefore proved that ${(g\circ f)}^{-1}(W)$ is open in $X$, as required.
For the second claim, suppose that $U$ is an open subset of $X$. Then ${\mathrm{id}}^{-1}(U)$ is just the same as $U$, and so is open. This proves that $\mathrm{id}$ is continuous. ∎
Let $X$ be a topological space, and let $Y$ be a subset of $X$. We declare that a subset $V\subseteq Y$ is open in $Y$ if there exists an open set $U$ of $X$ such that $V=U\cap Y$.
The above definition gives a topology on $Y$ (which we call the subspace topology).
The empty set can be written as the intersection of $Y$ with the open set $\mathrm{\varnothing}$ of $X$, so the empty set is open in $Y$. The full set $Y$ can be written as the intersection of $Y$ with the open set $X$ of $X$, so $Y$ is also open in $Y$.
Suppose we have a collection of sets ${V}_{i}\subseteq Y$ that are open in $Y$; we must show that the union ${V}^{*}={\bigcup}_{i}{V}_{i}$ is also open in $Y$. As each ${V}_{i}$ is open in $Y$, we can find open sets ${U}_{i}\subseteq X$ such that ${V}_{i}={U}_{i}\cap Y$. We are assuming that the axioms for a topology are satisfied by the open sets in $X$, so the union ${U}^{*}={\bigcup}_{i}{U}_{i}$ is again open in $X$. The set ${V}^{*}$ can be written as ${U}^{*}\cap Y$, so it is open as required.
Now suppose instead that we have a finite collection of sets ${V}_{1},\mathrm{\dots},{V}_{n}$ that are open in $Y$; we must show that the intersection ${V}^{\mathrm{\#}}={V}_{1}\cap \mathrm{\cdots}\cap {V}_{n}$ is also open in $Y$. As in (b), we can choose open subsets ${U}_{1},\mathrm{\dots},{U}_{n}$ with ${V}_{i}={U}_{i}\cap Y$. We are assuming that the axioms for a topology are satisfied by the open sets in $X$, so the intersection ${U}^{\mathrm{\#}}={\bigcap}_{i}{U}_{i}$ is again open in $X$. The set ${V}^{\mathrm{\#}}$ can be written as ${U}^{\mathrm{\#}}\cap Y$, so it is open as required.
∎
One very basic point about the subspace topology is as follows. We can define $i:Y\to X$ by $i(y)=y$ for all $y\in Y$. We call this the inclusion map.
If we give $Y$ the subspace topology, then the inclusion map $i\mathrm{:}Y\mathrm{\to}X$ is continuous.
Consider an open set $U\subseteq X$; we must show that ${i}^{-1}(U)$ is open with respect to the subspace topology on $Y$. But ${i}^{-1}(U)$ is just the same as $U\cap Y$, which is open by the definition of the subspace topology. ∎
We next deal with a slightly technical point. Suppose, for example, we want to discuss whether the function $f(x)=\mathrm{sin}(x)$ is continuous. We might want to regard this as a function from $\mathbb{R}$ to $\mathbb{R}$, or as a function from $\mathbb{R}$ to $[-1,1]$. We also might want to restrict the range of values of $x$, and consider $f$ as a function defined on $[0,2\pi ]$ or $[-\pi ,\pi ]$ or $[0,\mathrm{\infty})$ instead of all of $\mathbb{R}$. This leads us to worry about the following possibility: perhaps some of these versions of $f(x)$ are continuous, and some of them are not. It would then be a nightmare to keep track of everything. Fortunately, however, this nightmare does not arise: if we can check that $f$ is continuous as a map $\mathbb{R}\to [-1,1]$, then all other versions of $f$ are automatically continuous. This is the message of the following proposition.
Let $X$ and ${Y}_{\mathrm{1}}$ be topological spaces. Let ${X}_{\mathrm{0}}$ be a subset of $X$, and let $Y$ be a subset of ${Y}_{\mathrm{1}}$, with inclusion maps $i\mathrm{:}{X}_{\mathrm{0}}\mathrm{\to}X$ and $j\mathrm{:}Y\mathrm{\to}{Y}_{\mathrm{1}}$. Let $f$ be a continuous function from $X$ to $Y$. Let $\overline{f}\mathrm{=}j\mathrm{\circ}f\mathrm{\circ}i$ be the corresponding map ${X}_{\mathrm{0}}\mathrm{\to}{Y}_{\mathrm{1}}$, obtained by restricting the domain to ${X}_{\mathrm{0}}$ and enlarging the codomain to ${Y}_{\mathrm{1}}$, so we have a commutative diagram as follows:
Then $\overline{f}$ is also continuous.
There is also a partial converse for the above result.
Let $X$ and ${Y}_{\mathrm{1}}$ be topological spaces, and let $j\mathrm{:}Y\mathrm{\to}{Y}_{\mathrm{1}}$ be the inclusion of a subset with the subspace topology. Let $f$ be a function from $X$ to $Y$, and suppose that $j\mathrm{\circ}f$ is continuous (or equivalently: $f$ is continuous when regarded as a function $X\mathrm{\to}{Y}_{\mathrm{1}}$). Then $f$ is continuous (as a function $X\mathrm{\to}Y$).
Let $V$ be a subset of $Y$ that is open with respect to the subspace topology. We must show that ${f}^{-1}(V)$ is open in $X$. By the definition of the subspace topology, there must exist an open set ${V}_{1}\subseteq {Y}_{1}$ such that $V={V}_{1}\cap Y$. This can also be written as $V={j}^{-1}({V}_{1})$, so ${f}^{-1}(V)={f}^{-1}({j}^{-1}({V}_{1}))={(j\circ f)}^{-1}({V}_{1})$. We are given that $j\circ f:X\to {Y}_{1}$ is continuous and ${V}_{1}$ is open in ${Y}_{1}$ so ${(j\circ f)}^{-1}({V}_{1})$ is open in $X$. In other words, ${f}^{-1}(V)$ is open in $X$ as required. ∎
Let $Y$ be an open subset of $X$, and let $V$ be a subset of $Y$. Then $V$ is open with respect to the subspace topology on $Y$ iff $V$ is open with respect to the original topology on $X$.
First suppose that $V$ is open with respect to the subspace topology. By definition, this means that $V=U\cap Y$ for some subset $U\subseteq X$ that is open in $X$. Now both $U$ and $Y$ are open in $X$, so the intersection $V=U\cap Y$ is also open in $X$, as required.
Suppose instead that we start from the assumption that $V$ is open with respect to the original topology on $X$. If we just take $U=V$, we see that $U$ is open in $X$ and $V=U\cap Y$, so $V$ is open with respect to the subspace topology. ∎
Let $Y$ be an arbitrary subset of $X$, and let $G$ be a subset of $Y$. Then $G$ is closed (with respect to the subspace topology on $Y$) iff there exists a closed set $F\mathrm{\subseteq}X$ such that $G\mathrm{=}F\mathrm{\cap}Y$.
If $G$ is closed in $Y$, then the complement $V=Y\setminus G$ must be open in $Y$. This means by definition that there exists an open set $U\subseteq X$ such that $V=U\cap Y$. We can now put $F=X\setminus U$, which is a closed subset of $X$. We find that
$$F\cap Y=(X\setminus U)\cap Y=Y\setminus (U\cap Y)=Y\setminus V=G,$$ |
so $G$ has the form $F\cap Y$, as required. We leave the converse to the reader. ∎
Let $Y$ be a closed subset of $X$, and let $G$ be a subset of $Y$. Then $G$ is closed with respect to the subspace topology on $Y$ iff $G$ is closed with respect to the original topology on $X$.
Let $X$ be a topological space, and let $U$ be a subset of $X$. Suppose that for each $x\mathrm{\in}U$ we can find an open set $V$ such that $x\mathrm{\in}V$ and $V\mathrm{\subseteq}U$. Then $U$ itself is open.
For each point $x\in U$, choose a set ${V}_{x}$ as described, so ${V}_{x}$ is open and $x\in {V}_{x}$ and ${V}_{x}\subseteq U$. Put ${V}^{*}={\bigcup}_{x}{V}_{x}$. This is the union of a family of open sets, so it is open by the axioms for a topology. If we can prove that ${V}^{*}$ is the same as $U$, then we will be done. Each set ${V}_{x}$ is contained in $U$, and it follows that the union ${V}^{*}$ is also contained in $U$. On the other hand, for each $x\in U$ we have $x\in {V}_{x}\subseteq {V}^{*}$, so $U\subseteq {V}^{*}$. As $U\subseteq {V}^{*}$ and ${V}^{*}\subseteq U$ we have $U={V}^{*}$ as required. ∎
Let $f\mathrm{:}X\mathrm{\to}Y$ be a function between topological spaces. Suppose we have subsets ${U}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{U}_{n}\mathrm{\subseteq}X$ such that
Each set ${U}_{i}$ is open.
$X={U}_{1}\cup \mathrm{\cdots}\cup {U}_{n}$
For each $i$, the restricted map ${f}_{i}:{U}_{i}\to Y$ is continuous (with respect to the subspace topology on ${U}_{i}$).
Then $f$ is continuous.
Let $V$ be an open subset of $Y$; we must show that ${f}^{-1}(V)$ is open in $X$. Let ${f}_{i}:{U}_{i}\to Y$ be the restriction of $f$. As ${f}_{i}$ is continuous by assumption, we know that ${f}_{i}^{-1}(V)$ is open in ${U}_{i}$. By Lemma 3.30, this is the same as being open in $X$. Moreover, we see from the definitions that ${f}_{i}^{-1}(V)={f}^{-1}(V)\cap {U}_{i}$. As ${\bigcup}_{i}{U}_{i}=X$, we see that
$${f}^{-1}(V)=\bigcup _{i}({f}^{-1}(V)\cap {U}_{i})=\bigcup _{i}{f}_{i}^{-1}(V).$$ |
We have seen that each set ${f}_{i}^{-1}(V)$ is open, so the union of these sets is open, so ${f}^{-1}(V)$ is open as required. ∎
Again let $f\mathrm{:}X\mathrm{\to}Y$ be a function between topological spaces. Suppose we have subsets ${F}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{F}_{n}\mathrm{\subseteq}X$ such that
Each set ${F}_{i}$ is closed.
$X={F}_{1}\cup \mathrm{\cdots}\cup {F}_{n}$
For each $i$, the restricted map ${f}_{i}:{F}_{i}\to Y$ is continuous (with respect to the subspace topology on ${F}_{i}$).
Then $f$ is continuous.
By Proposition 3.23, it will be enough to show that ${f}^{-1}(G)$ is closed in $X$ whenever $G$ is closed in $Y$. After this preliminary step, the rest of the proof is essentially the same as for the previous proposition. ∎