Consider the metric space $H$ of uniformly continuous homeomorphisms (with uniformly continuous inverses) of some open bounded subset $G$ of $\mathbb C$, with the uniform metric $$d(f, g) = \sup_{z \in G} \left|f(z) - g(z)\right|.$$ It is clear that $(f, g) \mapsto f \circ g$ is a continuous map $H \times H \to H$ (and in fact this is due to the uniform metric, not the uniform continuity of any particular homeomorphism), but is inversion $f \mapsto f^{-1}$ a continuous map $H \to H$?

In other words, is it the case that whenever $(f_n) \to f$ uniformly on $G$, where $f$ and each $f_n$ is a uniformly continuous homeomorphism of $G$ with a uniformly continuous inverse, then $(f_n^{-1}) \to f^{-1}$ uniformly on $G$?

  • $\begingroup$ where are you stuck? $\endgroup$
    – Masacroso
    May 6, 2017 at 22:45
  • $\begingroup$ I'm trying to show that $(f_n^{-1} \circ f) \to id$ uniformly, which is equivalent. I expect this to be true since $(f_n) \to f$ uniformly, but I am having trouble putting this "additive" structure into use to prove the "multiplicative" structure that $f_n^{-1} \circ f$ becomes closer and closer to the identity. $\endgroup$
    – feralin
    May 6, 2017 at 22:49
  • $\begingroup$ IIRC this is not the case and the reason we usually take the metric $D(f,g) = d(f,g) + d(f^{-1}, g^{-1})$ using your $d$. $\endgroup$ May 6, 2017 at 22:58
  • $\begingroup$ I understand it is not usually the case, but in my question the space $G$ is bounded, which I hope may allow inversion to be continuous. Perhaps it still is not, though. $\endgroup$
    – feralin
    May 6, 2017 at 23:00
  • 1
    $\begingroup$ I don't have a reference for this specific fact. That there is an algebraic homomorphism follows from the fact that the completion of a metric space is a reflection arrow from the category of metric spaces with uniformly continuous maps to the full subcategory of complete metric spaces. It is not hard to see (from the density of a space in its completion) that this homomorphism is an isometry w.r.t. $d$, which implies that it must be a monomorphism. $\endgroup$ May 11, 2017 at 14:02

2 Answers 2


In fact $(H, \circ, d)$ will be a topological group whenever $(X, \rho)$ is a bounded metric space. (not necessarily totally bounded)

It is easy to see that $d(f \circ g, \operatorname{id}) \le d(f, \operatorname{id}) + d(g, \operatorname{id})$, hence composition is continuous at the identity.

Furthermore, it follows immediately from the surjectivity of $h$ that $d(f \circ h, g \circ h) = d(f, g)$, hence $d$ is right-invariant, so right translations are clearly continuous.

To show that $H$ is a topological group, it will suffice then to show that inversion is continuous.

To do this we take arbitrary $f \in H$ and $\epsilon > 0$. By the uniform continuity of $f^{-1}$ there is a $\delta > 0$ such that $\rho(f^{-1}(x), f^{-1}(y)) < \epsilon$ whenever $\rho(x, y) < \delta$. For any $g \in H$ such that $d(g, f) < \delta$, we have $$\rho(f^{-1}(g(x)), x) = \rho(f^{-1}(g(x)), f^{-1}(f(x))) < \epsilon/2$$ for all $x$, therefore $d(f^{-1} \circ g, \operatorname{id}) \le \epsilon/2 < \epsilon$. From the right-invariance of $d$ it follows that $$ d(f^{-1} \circ g, \operatorname{id}) = d(f^{-1} \circ g, g^{-1} \circ g) = d(f^{-1}, g^{-1}) $$ so we have $d(f^{-1}, g^{-1}) < \epsilon$, which shows that $f\mapsto f^{-1}$ is continuous.

Additional note: Without the assumption of uniform continuity, left translations may fail to be continuous, in which case inversion can not be continuous either. As an example take $X = (0,1)^2$ with $f(x,y) = (x^y, y)$ and $$ g_t(x, y) = \cases{ (2^tx, y) &when $x \le 4^{-t}$ \\ ((x-1)/(2^{-t} +1) &otherwise. } $$ Here $d(g_t, \operatorname{id}) = 2^{-t} - 4^{-t} < 2^-t$, so $g_t \to \operatorname{id}$ as $t \to \infty$.

On the other hand, taking $x = 4^{-t}, y = 1/t$, we can see that $d(f\circ g_t, f) \ge 2^{-1} - 4^{-1} \ge 1/4$, so $f\circ g_t$ does not converge uniformly to $f$, which shows that the left translation by $f$ is not continuous.

  • $\begingroup$ You gave a clear "no, inversion is not continuous" answer, so I am accepting your answer. $\endgroup$
    – feralin
    May 12, 2017 at 18:22

Let $K$ be a compact metric space, $H_K$ is the group of self-homeomorphisms $K\to K$ equipped with the topology of uniform convergence. I will prove that the inversion is continuous on $H_K$. If this were not the case, there would exist an (equicontinuous) sequence of homeomorphisms $h_n\to h$ such that $(h_n^{-1})$ does not converge to $h^{-1}$. By compactness, this means that (after passing to a subsequence in $(h_n)$) there exists $x\in K$, a sequence $(x_n)$ converging to $x$, while $h_n^{-1}(x_n)=y_n$ converges to $y\ne h^{-1}(x)$. Applying $h_n$ and $h$ and taking into account that $h_n\to h$, we obtain that $x_n=h_n(y_n)\to h(y)\ne x$. This contradicts the assumption that $x_n\to x$. qed

Now, apply this to the closure $K$ of your open and bounded subset $G\subset {\mathbb C}$.

Edit: Suppose that $f_n: K\to K$ is a sequence of continuous maps. What does it mean that $f_n$ does not converge to $f: K\to K$ (a continuous map) in uniform toplogy? This means (just by negating the definition of uniform convergence) that there exist $r>0$ such that for arbitrarily large $n$, there exist $x_n\in K$ such that $d(f_n(x_n), f(x_n))> r$. Now, by compactness, after extraction, we obtain $x_n\to x$, $f_n(x_n)\to y$ and $y$ has to be different from $f(x)$ since $d(y, f(x))\ge r$.

  • $\begingroup$ Sorry, I don't follow the important steps: the "by compactness" part and "we obtain that $x_n \to h(y) \neq x$. Could you please elaborate on these steps, the compactness consequence in particular? Thanks! $\endgroup$
    – feralin
    May 10, 2017 at 2:24
  • $\begingroup$ I think I see now. I didn't notice that you were using compactness to get two convergent sequences ($(x_n)$ and $(h_n^{-1}(x_n))$), so I was confused about how you knew that $(h_n^{-1}(x_n))$ actually converged. I understand now that you are really using compactness of $K^2$ to get a subsequence of the zippered sequence $(x_n, h_n^{-1}(x_n))$ in $K^2$, which is what you need. $\endgroup$
    – feralin
    May 10, 2017 at 4:14
  • $\begingroup$ @feralin: Right. $\endgroup$ May 10, 2017 at 4:32
  • $\begingroup$ Moishe, sorry, I also don't see how to apply your answer to homeomorphisms of $G$. You say to apply the fact that inversion is continuous on self-homeomorphisms of $\bar G$ (equipped with uniform convergence topology) to deduce that inversion is continuous on self-homeomorphisms of $G$, also with the uniform convergence topology, but I don't see how to do this. Could you clarify please? $\endgroup$
    – feralin
    May 10, 2017 at 22:59
  • $\begingroup$ I've tried to prove that deduction, but I keep running into the problem that a homeomorphism of $G$ need not extend continuously to a homeomorphism of $\bar G$. $\endgroup$
    – feralin
    May 10, 2017 at 23:01

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