Why is $\text{Homeo}_p(X)$, the space of homeomorphisms of $X$ with topology of pointwise convergence and operation of composition not a topological group for $X = \mathbb{R}^2$?

This question was asked before and got an answer here but without proof.

It's clearly a quasi-topological group, the author claims composition is not jointly continuous. I found somewhere it's a topological group for $X = \mathbb{R}$ for example, so the author might be wrong (not sure).

I'm looking for a proof.

  • 2
    $\begingroup$ @TheSilverDoe you're saying this like it's some kind of homework $\endgroup$
    – Jakobian
    Jul 11, 2023 at 17:26
  • 3
    $\begingroup$ Homework or not, this makes no difference, a question has to meet the quality standards (and you are here for quite a long time, you should know this...). $\endgroup$ Jul 11, 2023 at 17:28
  • 4
    $\begingroup$ Why are you being so hostile in the comments? You're literally just asking a question with no effort whatsoever to try and answer it yourself. That is a textbook low quality question, homework or not. $\endgroup$
    – Lorago
    Jul 11, 2023 at 18:06
  • 4
    $\begingroup$ @Lorago I would disagree with you. I think OP provided context (they read the statement in another answer), and questions of the type "wondering out loud" are often well received and don't benefit from additional context or "attempts". $\endgroup$
    – s.harp
    Jul 11, 2023 at 20:11
  • 3
    $\begingroup$ @s.harp The question looked very different two hours ago, when the comments were left. $\endgroup$
    – Xander Henderson
    Jul 11, 2023 at 20:16

2 Answers 2


Let $g_n(x)= x + (1/n,0)$. This converges pointwise to the identity map.

Let $f_n$ be a homeomorphism that maps the point $(1/n,0)$ to $(1/n,1)$ and is the identity outside of $U_n:=B_{2^{-n}}(\{1/n\}\times[0,1])$. These kind of things can be built by multiplying the vector field $(0,1)$ with an appropriate function that is supported in $U_n$ and equal to $1$ on the interval $\{1/n\}\times[0,1]$ and then taking the flow.

Then for any $x\in\Bbb R^2$ there is an $N$ so that $x\notin U_n$ for $n>N$, and then $f_n$ converges pointwise to the identity map.

But $f_n(g_n((0,0)))=(1/n,1)$ converges to $(0,1)$ and not $(0,0)$.

Basically whats happening is that $f_n$ is always pushing some point close to the origin far away, but as far as pointwise convergence can tell the set on which this happens is being "shuffled out of existence". The function $g_n$ is chosen so that the origin is mapped to this nearby point.


More generally, suppose $X$ is a topological space which is strongly homogeneous in the sense that any finite sequence of distinct points of $X$ can be sent to any other such finite sequence by a homeomorphism $X\to X$. Then as long as every nonempty open subset of $X$ is infinite and there exists an nonempty proper open subset $U_0\subset X$, $\operatorname{Homeo}_p(X)$ is not a topological group. (In particular, this applies to any connected manifold of dimension greater than $1$; see here for instance.)

To prove this, pick a point $x\in U_0$ let $U\subseteq \operatorname{Homeo}_p(X)$ be the set of homeomorphisms $f$ such that $f(x)\in U_0$. This $U$ is an open neighborhood of $1\in\operatorname{Homeo}_p(X)$. If composition were continuous, there would then be some basic open neighborhood $V\subseteq \operatorname{Homeo}_p(X)\times \operatorname{Homeo}_p(X)$ of $(1,1)$ such that $f\circ g\in U$ for all $(f,g)\in V$. Such a $V$ can be taken to have the following form: for some points $x_1,\dots,x_n\in X$ and open sets $V_1,\dots,V_n$ with $x_i\in V_i$, $V$ is the set of $(f,g)$ such that $f(x_i),g(x_i)\in V_i$ for each $i$. We may further assume that $x_1=x$. Now pick a point $y\in V_1$ which is different from all the $x_i$ and a point $z\in X\setminus U_0$. By our homogeneity assumption, there exists $g\in \operatorname{Homeo}_p(X)$ such that $g(x)=y$ and $g(x_i)$ is some chosen point in $V_i$ for each $i>1$. There also exists $f\in \operatorname{Homeo}_p(X)$ such that $f(y)=z$ and $f(x_i)$ is some chosen point in $V_i$ for each $i$. Then $(f,g)\in V$, but $f(g(x))=z\not\in U_0$ so $f\circ g\not\in U$ This contradicts our choice of $V$, so composition cannot be continuous.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .