Why is the number of connected components invariant under homeomorphisms? I know that connectedness, as well as path connectedness, are properties conserved through homeomorphisms. But why is this really meaningful? For example, consider the set

$$A:=\{ (x,y) \in \mathbb{R}^2 : xy = 0\}.$$

I know this set is not a 1-manifold since for $A \setminus \{(0,0)\}$ any neighborhood of $(0,0)$ has 4 connected components and an open interval in $\mathbb{R}$ without a point has just 2. However, I think there could be some further details to take care rather than just comparing "4 against 2".

Could it be maybe possible to show this using the zero homology $H_0$?

Many thanks

  • 3
    $\begingroup$ I'm not sure if I understand the question. Connected components, cut points, and such are invariants of homeomorphisms, but they are not complete invariants. The best they can do is sometimes tell whether certain spaces are non-homeomorphic. $\endgroup$ – Kyle Miller Mar 5 at 1:52

A component $C$ of $X$ is a maximal connected subset of $X$: $C$ is connected and if $C \subseteq C' \subseteq X$ and $C'$ is connected then $C'=C$.

It is standard that any two distinct components of $X$ are disjoint. So they form a partition of the space: to see they cover $X$ note that for any $x \in X$, $C_x := \bigcup \{ C \subseteq X: C \text{ connected }, x \in C\}$ is connected (as a (non-empty, as the connected set $\{x\}$ is one of the $C$ we use) union of intersecting connected sets) and is clearly maximal, so $C_x$ is a component of $X$ that contains $x$.

Let $\mathscr{C}(X)$ be the set of components of $X$, and let $h: X \to Y$ be a homeomorphism.

Then the induced map $\bar{h}: \mathscr{C}(X) \to \mathscr{C}(Y)$ can be defined by $\bar{h}(C)=h[C] \subseteq Y$ (just the image set under $h$)

$\bar{h}(C)$ is connected as the continuous image of a connected set. And if $\bar{h}(C) = h[C] \subseteq C' \subseteq Y$ and $C'$ is connected, then $C\subseteq h^{-1}[C']$ and the latter set is connected (as $h^{-1}$ is continuous too) and so maximality in $X$ tells us that $C= h^{-1}[C']$ or $\bar{h}(C) = C'$ (applying the bijective $h$ to both sides). So indeed $\bar{h}(C) \in \mathscr{C}(Y)$.

$\bar{h}$ has the same induced map based on $h^{-1}$ as its inverse so we have a bijection between $\mathscr{C}(X)$ and $\mathscr{C}(Y)$ so in particular the number (cardinality) of their components, call it $\textrm{nc}(X)$, say, is the same.

So $$X \simeq Y \implies \operatorname{nc}(X) = \operatorname{nc}(Y)$$ which can be used in its contrapositive form

$$\operatorname{nc}(X) \neq \operatorname{nc}(Y) \implies X \not\simeq Y$$

to show spaces are non-homeomorphic.

In your argument you also implicitly use the following fact:

If $X \simeq Y$ by a homeomorphism $h$, then $X\setminus \{p\} \simeq Y \setminus \{h(p)\}$ for any $p \in X$.

the proof of which is trivial.


Let $C(x,S)$ be the connected component of $S$ that contains $x$.
Assume $$h:X \to Y$$ is an homeomorphism. Show $h[C(x,X)] = C(h(x),Y)$ for all $x \in X$. Also show if $C(h(x),Y) = C(h(y),Y)$, then $C(x,X) = C(y,Y)$. Thus $h(\cdot)$ is an injection from the components of $X$ to those of $Y$.

As $h^{-1}: Y \to X$ is a homeomorphism, likewise there is an injection from the components of Y to those of X. Thus by the Cantor–Bernstein theorem, the components of $X$ is equinumerous to the components of Y.

  • $\begingroup$ Sir. Let me check if I follow your argument. You are trying to perform a bijection between the connected components of the sets $X,Y$. So in this case from the Cantor-Bernstein you follow that $X,Y$ are equinumerous in terms of number of connected components. Rigth? $\endgroup$ – rarc Mar 5 at 9:08
  • $\begingroup$ Thar is correct. @rarc $\endgroup$ – William Elliot Mar 5 at 11:45

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