topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
continuous metric space valued function on compact metric space is uniformly continuous
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Analysis Theorems
An embedding of topological spaces is a continuous function which is a homeomorphism onto its image.
(embedding of topological spaces)
Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A continuous function $f \;\colon\; X \longrightarrow Y$ is called an embedding if
in its image factorization
with the image $f(X) \hookrightarrow Y$ equipped with the subspace topology, then $X \to f(X)$ is a homeomorphism.
(open/closed continuous injections are embeddings)
A continuous function $f \colon (X, \tau_X) \to (Y,\tau_Y)$ which is
an open map or a closed map
is an embedding (def. 1).
This is called a closed embedding if the image $f(X) \subset Y$ is a closed subset.
If $f$ is injective, then the map onto its image $X \to f(X) \subset Y$ is a bijection. Moreover, it is still continuous with respect to the subspace topology on $f(X)$. Now a bijective continuous function is a homeomorphism precisely if it is an open map or a closed map (by this prop.). But the image projection of $f$ has this property, respectively, if $f$ does (by this prop.).
Embeddings of topological spaces are precisely the regular monomorphisms in the category Top of all topological spaces.
For proof see at Top this proposition.
In $Top$, the pushout $j$ of a (closed/open) embedding $i$ along any continuous map $f$,
is again a (closed/open) embedding.
For proof see at subspace topology here.
(injective proper maps to locally compact spaces are equivalently the closed embeddings)
Let
$X$ be a topological space
$f \colon X \to Y$ be a continuous function.
Then the following are equivalent
$f$ is an injective proper map,
$f$ is a closed embedding (def. 1).
In one direction, if $f$ is an injective proper map, then since proper maps to locally compact spaces are closed, it follows that $f$ is also closed map. The claim then follows since closed injections are embeddings, and since the image of a closed map is closed.
Conversely, if $f$ is a closed embedding, we only need to show that the embedding map is proper. So for $C \subset Y$ a compact subspace, we need to show that the pre-image $f^{-1}(C) \subset X$ is also compact. But since $f$ is an injection (being an embedding), that pre-image is just the intersection $f^{-1}(C) \simeq C \cap f(X)$.
To see that this is compact, let $\{V_i \subset X\}_{i \in I}$ be an open cover of the subspace $C \cap f(X)$, hence, by the nature of the subspace topology, let $\{U_i \subset Y\}_{i \in I}$ be a set of open subsets of $Y$, which cover $C \cap f(X) \subset Y$ and with $V_i$ the restriction of $U_i$ to $C \cap f(X)$. Now since $f(X) \subset Y$ is closed by assumption, it follows that the complement $Y \setminus f(X)$ is open and hence that
is an open cover of $C \subset Y$. By compactness of $C$ this has a finite subcover. Since restricting that finite subcover back to $C \cap f(X)$ makes the potential element $Y \setminus f(X)$ disappear, this restriction is a finite subcover of $\{V_i \subset C \cap f(X)\}$. This shows that $C \cap f(X)$ is compact.