Homeomorphism / Open Map / Close Map / Embedding

Yao Yao on October 7, 2018


Quoted from Wikipedia: Homeomorphism:

A function $f:X \to Y$ between two topological spaces $(X, \tau_{X})$ and $(Y,\tau_{Y})$ is called a homeomorphism if it has the following properties:

  • $f$ is a bijection,
  • $f$ is continuous,
  • and the inverse function $f^{-1}$ is continuous.

A homeomorphism is sometimes called bicontinuous. If such a homeomorphism exists, we say $(X, \tau_{X})$ and $(Y,\tau_{Y})$ are homeomorphic.

A self-homeomorphism is a homeomorphism of a topological space and itself.

The homeomorphisms form an equivalence relation on the class of all topological spaces. The resulting equivalence classes are called homeomorphism classes.


  • $f(x) = \frac{1}{a - x} + \frac{1}{b - x}$ for $a < b$ is a homeomorphism between open interval $(a, b)$ and $\mathbb{R}$.
  • The graph of a differentiable function is homeomorphic to the domain of the function.
    • 假设 $f$ 是 differentiable,那么 $g(x) = (x, f(x))$ 是 homeomorphism

Open Map / Close Map

An open map is a function between two topological spaces which maps open sets to open sets. That is, a function $f:X \to Y$ is open if $\forall$ open set $U \subset X$, the image $f(U)$ is open in $Y$.

Likewise, a closed map is a function which maps closed sets to closed sets.

A map may be open, closed, both, or neither.

Proposition: A continuous bijection $f$ that is closed is also open (and vice versa).

Proof: Suppose $f$ is a closed, continuous bijection, and $V$ is open.

Then the $V^c$ is closed, and therefore the image $f(V^c)$ is closed, since $f$ is closed.

But bijectivity means that $f(V^c)=(f(V))^c$. This means that $f(V)$ is open, being the complement of a closed set. Thus, $f$ is an open map as well.

The proof that an open, continuous bijection is also closed is completely analoguous. $\blacksquare$

Proposition: A bijective continuous map $f$ is a homeomorphism $\iff$ $f$ is open, or equivalently, is closed.

Proof: $f^{-1}$ must exist because $f$ is a bijection.

So equivalently we just need to prove: for a bijective continuous map $f$, its inverse $f^{-1}$ is continuous $\iff$ $f$ is open, or equivalently, is closed.

  • 这里我们需要 topological space 上 continuous function 的定义:
    • A function $f:X \to Y$ between two topological spaces $(X, \tau_{X})$ and $(Y,\tau_{Y})$ is continuous if $\forall$ open set $V \subseteq Y$, the inverse image $f^{-1}(V)=\lbrace x \in X \mid f(x) \in V \rbrace \subseteq X$ is open

从这个定义来看,这个 proof 是 trivial 的。$\blacksquare$


Quoted from Wikipedia: Embedding:

In general topology, an embedding is a homeomorphism onto its image.

More explicitly, an injective continuous function $f:X \to Y$ between two topological spaces $(X, \tau_{X})$ and $(Y,\tau_{Y})$ is a topological embedding if $f: X \to f(X)$ is a homeomorphism between $X$ and $f(X)$ (where $f(X)$ carries the subspace topology inherited from $Y$).

  • 注意:如果 $f$ 是 injective 的,那么 $f: X \to f(X)$ 必然是 surjective 的,进而必然是 bijective 的
  • A homeomorphism is natually an embedding;反之不成立

Intuitively then, the embedding $f:X \to Y$ lets us treat $X$ as a subspace of $Y$.

  • 我觉得这句话才是理解 embedding 的精华所在
  • 因为 homeomorphism 定义了 topological space 的一种等价关系,用 $\cong$ 表示的话,我们可以写成 $X \cong f(X) \subset Y$,这个式子就可以理解为 “把 $X$ embed 到 $Y$”

Proposition: Every function that is injective, continuous and either open or closed is an embedding.

However there are also embeddings which are neither open nor closed. It happens if the image $f(X)$ is neither an open set nor a closed set in $Y$.

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