Topological space

Let $X$ be a set. A topology over $X$ is some subset $\tau\subseteq \mathcal{P}(X)$ such that:

$\emptyset, X\in \tau$ .
$\{U_i\}_{i\in I}\subseteq \tau \Longrightarrow \bigcup_{i\in I} U_i\in\tau$ .
$\{U_i\}_{i=1,\ldots,n}\subseteq \tau \Longrightarrow \bigcap_{i=1}^{n} U_i\in\tau$ .

The subsets $A\in\tau$ are called the open subsets of the topology.

A topological space is a set $X$ endowed with some specific topology $\tau$ . Despite a topological space is technically a pair $(X, \tau)$ , we might as well refer to the "topological space $X$ " whenever the topology is understood, e.g. the topological space $\R$ (whose open sets are the clasical open sets of real analysis).

Initial topologies and embeddings

Definition.(Initial topologies)

Let $f\colon X\to Y$ be a map. Suppose $Y$ is endowed with some topology. We define the initial topology on $X$ with respect to $f$ as

\tau = \{f^{-1}(U)\mid U\in \tau\}

This is, the coarsest topology on $X$ such that the map $f\colon X\to Y$ is continuous.

We can suppose that $X$ and $Y$ are already topological spaces and ask ourselves which conditions on $f$ guarantee that the map $f\colon X\to Y$ is "initial", i.e., $X$ has the initial topology. By definition, if there're some $x_1, x_2\in X$ s.t. $f(x_1) = f(x_2)$ then the open neighborhoods of $x_1$ and $x_2$ are exactly the same. Thus, $x_1$ and $x_2$ are topologically indistinguishable (see Kolmogorov axiom). Thus, we can "collapse" the topologically equivalent points in the domain and get an initial and injective function $f$ . Being initial thus becomes equivalent to the fact that $f$ is a homeomorphism of $X$ and $f(X)$ , which is usually called being an embedding.

Definition.(Embeddings)

Let $f\colon X\to Y$ be a continuous map between topological spaces. We say that $f$ is an embedding, or, equivalently, that $X$ is an (embedded) subspace of $Y$ , if $F: X\to Y$ is an homeomorphism between $X$ and $f(X)$ . Equivalently, a map $f$ is an embedding if and only if $f$ is injective and $X$ has the initial topology with respect to $f$ .

For example, let us consider the map $f: [0, 1)\to \R^2$ given by $f(t) = (\cos{2\pi t}, \sin{2\pi t})$ . If we take the usual topology if $[0, 1]$ , then this map is not an embedding since the open set $[0, \varepsilon)\subseteq [0, 1)$ cannot be the preimage of any open set in $\R^2$ . This occurs as $[0, 1)$ is definitely not homeomorphic to the subspace $S^1\subseteq \R^2$ .

If we take the mapping $f(t) = (\cos{\pi t}, \sin{\pi t})$ instead, this function is an embedding, as a semicircumference is homeomorphic to a segment.

Otherwise, we could instead choose the initial topology on $[0, 1)$ given by the former mapping. With this topology $[0, \varepsilon)$ is not an open set, as in fact every neighborhood of $0$ must contain $1-\varepsilon$ for some $\varepsilon$ . This is like giving the " $S^1$ topology" to the semiopen interval, which entails "gluing" the endpoints together.

Hausdorff axiom

Our current definition of topological space has some drawbacks. There're some "intuitive" facts that do not hold anymore for arbitrary topological spaces:

A singleton $\{x\}$ does not need to be a closed subset.
A sequence $(x_n)_{n\ge 1}$ can have more than one limit.

(Prove that if every sequence has a unique limit then the set $\{x\}$ must be closed.)

There is an additional axiom which allows us to prove the previous two facts for every topological space that satisfies it.

Definition.(Hausdorff axiom)

Let $X$ be a topological space. We say that $X$ satisfies the Hausdorff axiom (also called the $T_2$ axiom) if for every $x, y\in X$ there're open sets $U$ , $V$ such that $x\in U$ , $y\in V$ and $U\cap V = \emptyset$ .

Kolmogorov axiom

We now study an axiom which is strictly weaker than the Hausdorff axiom.

Definition.(Kolmogorov)

Let $X$ be a topological space. We say that $X$ satisfies the Kolmogorov axiom (also called the $T_0$ axiom) if for every $x, y\in X$ there is an open set $U$ such that $x\in U$ and $y\not\in U$ or $x\not\in U$ and $y\in U$ .

It is remarkable that all spaces satisfying this axiom can be characterized in a very specific way.

Compactness

Definition.(compact spaces)

Let $X$ be a topological space. We say that $X$ is compact if every open cover of $X$ admits a finite subcover.

Theorem.(equivalences of compactness)

The following are equivalent definitions of compactness over an arbitrary topological space $X$ :

every open cover of $X$ admits a finite subcover.
every family of open sets with the FIP has non-empty intersection.
every net has a convergent subnet.
the function $f: X\times Z\to X$ is closed for every space $Z$ .

The equivalence of 1. and 2. is evident.

To prove that 1. implies 3., we consider an arbitrary net $(x_\lambda)_{\lambda\in \Lambda}$ . Suppose the net doesn't have a convergent subnet. Then, for every $x\in X$ , there is some open neighbourhood $U_x$ such that no subnet is eventually contained in $U_x$ . This means that the set $A_x = \{\lambda\in\Lambda\mid x_\lambda\in U_x\}$ is not cofinal, which means that there is some $\lambda_x\in \Lambda$ such that there is no $\alpha\in A_x$ with $\lambda_x\le \alpha$ . As every open cover has a finite subcover, there is a finite set $x_1, x_2, \ldots, x_n$ such that their corresponding open sets cover all $X$ . If we take $\lambda_0 \ge \lambda_{x_1}, \ldots, \lambda_{x_n}$ , then there cannot be any $\lambda\ge \lambda_0$ in $\Lambda$ . As $\Lambda$ is directed, this means that $\lambda_0$ is an upper bound, and therefore $x_{\lambda_0}$ must be an accumulation point of the net, a contradiction.

To prove that 3. implies 2., we consider an arbitrary family $(F_i)_{i\in\mathcal{I}}$ with the FIP. Thus, for every finite $\mathcal{J}\subseteq\mathcal{I}$ we have some $x_\mathcal{J}\in \bigcap_{j\in\mathcal{J}}{F_j}$ . This gives a net $(x_{\mathcal{J}})$ directed by the poset of finite subsets of $\mathcal{I}$ , which must have a convergent subnet. Let $x_0$ be the limit point, $x_0\in U$ an arbitrary neighbourhood and $i\in \mathcal{I}$ an arbitrary index. Because every subnet is cofinal, there is some $\mathcal{J}\supseteq \{i\}$ such that $x_\mathcal{J}\in U$ , which means that $F_i\cap U\neq \varnothing$ . As this assertion is true for every $U$ , then $x_0$ is a limit point of $F_i$ , which is closed, and thus $x_0\in F_i$ . As this is true for every $i\in\mathcal{I}$ , we have that $x_0\in\bigcap_{i\in\mathcal{I}}{F_i}$ , as desired.

Definition.(proper maps)

A continuous map $f\colon X\to Y$ is proper if it is closed and its fibers are compact.

Theorem.

The following definitions of proper map are equivalent:

it is closed and its fibers are compact.
preimage of compact sets is compact.

Theorem.

Composition of proper maps is proper.

We can actually put the concept of compactness into a even more general setting, which involves using pullbacks and universally closed maps.

Definition.(fibered products)

Let $X$ , $Y$ , $Z$ be topological spaces and $f: X\to Z$ , $g: Y\to Z$ continuous maps. Then, the fibered product of $X$ and $Y$ through $f$ and $g$ is the space

X\times Y = \left\{(x, y)\in X\times Y\mid f(x) = g(y)\right\}

with the subspace topology.

Definition.(pullbacks)

Let $X$ , $Y$ , $Z$ be topological spaces and $f: X\to Z$ , $g: Y\to Z$ continuous maps. Then, the pullback of $X$ and $Y$ through $f$ and $g$ is the unique space $W$ , endowed with projections $\overline{f}\colon W\to Y$ and $\overline{g}\colon W\to X$ such that $f\overline{g} = g\overline{f}$ with the following universal property: for every other space $W'$ endowed with projections $f'\colon W'\to Y$ and $g'\colon W'\to X$ such that $fg' = gf'$ , there is a unique continuous function $\varphi\colon W'\to W$ such that $f' = \varphi \overline{f}$ and $g' = \varphi \overline{g}$ .

This is, given an "incomplete" commutative square, the pullback is the missing corner which not only makes the diagram commute, but it is also the closest object which makes the diagram commute.

Definition.(universally closed maps)

A continuous map $f\colon X\to Y$ is universally closed if for every topological space $Z$ , the continuous map $f\times\text{id}_Z\colon X\times Z\to Y\times Z$ is closed.

Theorem.

The following definitions of universally closed map are equivalent:

for every topological space $Z$ , the continuous map $f\times\text{id}_Z\colon X\times Z\to Y\times Z$ is closed.
for every topological space $Z$ , if $W$ is the pullback of $X$ and $Z$ through $Y$ with projection $\overline{f}\colon W\to Z$ , then $\overline{f}$ is closed.

The implication from 2. to 1. is trivial.

Theorem.

Given a continuous function $f\colon X\to Y$ , then $f$ is proper if and only if it is universally closed.