I realized that, for any pair of non-homeomorphic topological spaces that I know of, those three invariants are usually sufficient to prove that the two spaces are not the same. So, for example:

  • the interval $[0,1]$ and the real line $\mathbb{R}$ are both connected and their fundamental group is the trivial group, but the former is compact while the latter isn't;
  • the interval $[0,1]$ and the union of intervals $[0,1]\cup[2,3]$ are both compact sets and their fundamental group is the same for each of their connected components, but the former is connected while the latter isn't;
  • the circumference $S^1$ and the spherical surface $S^2$ are both connected and compact, but $\pi(S^1) = \mathbb{Z}$ while $\pi(S^2) = \{1\}$.

But I was wondering: can I find two topological spaces that are both compact (or not compact) and connected (or disconnected)—and that also have the same fundamental group—despite not being homeomorphic? If yes, what other topological invariant(s) could help us understand they're not the same space?

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    $\begingroup$ Cardinality is technically a topological invariant, as is whether or not a space is nonempty. :) $\endgroup$ Commented Oct 25, 2018 at 19:08
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    $\begingroup$ Its also worth noting that a topological invariant is essentially just a collection of topological spaces that is closed under homeomorphisms. The invariant is simply whether or not a space is included in that collection. $\endgroup$ Commented Oct 25, 2018 at 19:25
  • $\begingroup$ There is a large and likely growing collection of infinite cardinals associated with any topological space, called topological cardinals, or topological cardinal invariants, which are invariant under homeomorphisms. Some of their names are weight, density, cellularity, extent, Lindelof number. There is a lot about them in various parts of General Topology, by R. Engelking... I dk if this has anything to do with fundamental groups. $\endgroup$ Commented Oct 25, 2018 at 22:27
  • $\begingroup$ see this post. $\endgroup$
    – C.F.G
    Commented Oct 1, 2020 at 16:42

2 Answers 2


Without straying into the realms of algebraic topology a very natural invariant to consider is dimension. The primary problem was (and still is) how dimension should be defined. There are several different definitions for dimension. The three most popular definitions are the small and large inductive dimensions, and the covering dimension. I will give the definitions of these below, but Engelking's "Dimension Theory" (which may or may not be available online wink wink) is an excellent introduction to the subject.

Definition Given a regular space $X$ the small inductive dimension of $X$, denoted $ind(X)$ is defined inductively by defining $ind(X)=-1$ if and only if $X=\emptyset$. Then for a natural number $n$ we say that $ind(X)\leq n$ if for every point $x\in X$ and open neighbourhood $U\subseteq X$ of $x$ there is an open neighbourhood $V\subseteq X$ of $x$ such that $x\in V\subseteq U$ and $ind(\partial V)\leq n-1$ where $\partial V$ is the boundary of $V$ in $X$. We say that $ind(X)=n$ if $ind(X)\leq n$ is true and $ind(X)\leq n-1$ is not. We say that $ind(X)=\infty$ if $ind(X)\leq n$ is not true for every $n$.

If in the above definition $X$ is a normal space and the point $x$ is replaced with a closed subset $C$ of $X$, then one gets the definition of the large inductive dimension of $X$, denoted $Ind(X)$.

NOTE The large and small inductive dimensions coincide on separable metric spaces. Most importantly $ind(\mathbb{R}^{n})=Ind(\mathbb{R}^{n})=n$ for each $n$. However, they do not coincide for nonseparable metric spaces. Roy Prabir famously constructed a complete metric space whose small inductive dimension is $0$, but whose large inductive dimension is $1$. If my memory serves correctly that gap has not been widened for nonseparable metric spaces.

The covering dimension as defined by Cech is as follows.

Definition A finite open cover $U_{1},U_{2},\ldots,U_{n}$ of a normal space $X$ is said to have order $n$ if every point of $x$ belongs to at most $n$ $U_{i}$. An open cover $\mathcal{U}$ of $X$ is refined by an open cover $\mathcal{V}$ of $X$ if for all $V\in\mathcal{V}$ there is some $U\in\mathcal{U}$ such that $V\subseteq U$.

Definition Let $X$ be a normal space. The (Cech) covering dimension of $X$, commonly denoted by $dim(X)$, is defined to be $-1$ iff $X=\emptyset$. Otherwise it is the least natural number $n$ such that every finite open cover $\mathcal{U}$ of $X$ is refined by a finite open cover $\mathcal{V}$ of order at most $n+1$. As always $dim(X)=\infty$ if $dim(X)\leq n$ is not true for each $n$.

The covering dimension so defined coincides with the inductive dimensions on separable metric spaces. As could be expected, all coincidences go out the window when you drop separability. However, $dim(X)=Ind(X)$ for all metrizable spaces. It is also worth noting that for paracompact spaces this definition of the covering definition is equivalent to the, perhaps more common, notion of Lebesgue covering dimension.

  1. Homology groups, related to homotopy groups (the abelianization of the fundamental group (the first homotopy group) is the first homology group).
  2. Cohomology groups, related to homology groups via Poincare duality.

If your topological space is a low-dimensional manifold or a knot, then there are plenty of invariants.


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