# Is the space $\text{Map}(S^1,S^1)$ of continuous maps on $S^1$ compact? (Compact-open topology)

Is the space $$\text{Map}(S^1,S^1)$$ of continuous maps $$S^1\to S^1$$ compact? Here $$\text{Map}(S^1,S^1)$$ has the compact-open topology.

I'm not too savvy with the compact-open topology so I'm not sure where to really begin. I also tried searching this question and couldn't find it on here. I know that $$\text{Map}(S^1,S^1)$$ has a prebasis (subbasis) given by all $$S(K,U):=\{f:S^1\to S^1\mid f(K)\subseteq U\}$$ where $$K\subseteq S^1$$ is compact and $$U\subseteq S^1$$ is open. But I don't know how to relate this to the compactness of $$\text{Map}(S^1,S^1)$$. Any hints?

• Have you tried leveraging the fact that $S^1$ is itself compact? Commented May 29, 2021 at 18:56
• Kinda: if we start with an open cover $\text{Map}(S^1,S^1)=\bigcup_{a\in I}U_a$, each $U_a$ can be written as a union of a finite intersection of the $S(K,U)$. Now each constant map is continuous, so there has to be enough $S(K,U)$ belonging to $\bigcup_{a\in I}U_a$ so that the codomain $S^1$ gets covered by open sets $U$, from which we could extract a finite subcover via the compactness of $S^1$. But then I don't see how to build a finite subcover of $\bigcup_{a\in I}U_a$ out of this.. Commented May 29, 2021 at 19:15
• By Arzela-Ascoli it suffices to prove equicontinuity and that $\{f(y) : f\in \text{Maps}(S^1, S^1)\}$ is relatively compact for all $y$ Commented May 29, 2021 at 19:18
• After googling Arzela-Ascoli, I see how that approach could work, but I was hoping for a more direct, less analytical approach... Commented May 29, 2021 at 19:26

The space $$\operatorname{Map}(S^1,S^1)$$ is not compact. To see this, we can use the following well-known theorems:

1. If $$X$$ is a compact Hausdorff space and $$(Y,d)$$ is a metric space, then the compact-open topology on $$\operatorname{Map}(X,Y)$$ agrees with the metric topology induced by the supremum metric $$d(f,g) = \sup\{d(f(x),g(x)) \mid x \in X \}$$.
See for example Topology on the set of Continuous Functions from a Compact to a Metric Space.

2. Each compact metric space is sequentially compact (which means that each sequence has a convergent subsequence).
See fort example In a metric space, compact implies sequentially compact.

Now assume that $$\operatorname{Map}(S^1,S^1)$$ is compact. Consider the sequence $$f_n(z) = e^{i\lvert \operatorname{Re}(z) \rvert^n}$$ (recall $$S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\\$$). It must have a convergent subsequence (with respect to the supremum metric) $$(f_{n_k})$$ with limit $$f \in \operatorname{Map}(S^1,S^1)$$. Clearly $$f$$ is also the pointwise limit of this sequence.

For $$z \ne \pm 1$$ we have $$\lvert \operatorname{Re}(z) \rvert < 1$$, hence $$\lim_{k \to \infty}\lvert \operatorname{Re}(z) \rvert^{n_k} = 0$$ and therefore $$f(z) = \lim_{k \to \infty}f_{n_k}(z) = e^{i0} = e^0 = 1$$. For $$z = \pm 1$$ we have $$\lvert \operatorname{Re}(z) \rvert = 1$$, hence $$\lvert \operatorname{Re}(z) \rvert^{n_k} = 1$$ for all $$k$$ and therefore $$f(\pm1) = e^{i1} = e^i \ne 1$$. This shows that $$f$$ is not continuous which contradicts $$f \in \operatorname{Map}(S^1,S^1)$$.

Therefore $$\operatorname{Map}(S^1,S^1)$$ is not compact.

Remark:

The compactness of a function space $$\operatorname{Map}(X,Y)$$ enforces serious requirements concerning $$X, Y$$. Here are two fairly obvious results:

1. If $$\operatorname{Map}(X,Y)$$ is compact and $$X \ne \emptyset$$, then $$Y$$ is compact.
In fact, let $$\mathfrak U$$ be an open cover of $$Y$$. Pick $$x \in X$$. Then the sets $$S(\{x\},U)$$ with $$U \in \mathfrak U$$ form an open cover of $$C_{co}(X,Y)$$. If this space is compact, then there exist finitely many $$U_i \in \mathfrak U$$ such that the $$S(\{x\},U_i)$$ cover $$C_{co}(X,Y)$$. Thus for each $$y \in Y$$ the constant functions $$c_y : X \to Y$$ with value $$y$$ is contained in some $$S(\{x\},U_i)$$. But then $$y = c_y(x) \in U_i$$. Therefore the $$U_i$$ cover $$Y$$.

2. If $$Y$$ is compact and $$X$$ is discrete, then $$\operatorname{Map}(X,Y)$$ is compact.
In fact, for a discrete $$X$$ we have $$\operatorname{Map}(X,Y) \approx \prod_{x \in X} \operatorname{Map}(\{x\},Y)$$ which is a product of compact spaces since $$\operatorname{Map}(\{x\},Y) \approx Y$$.

Exercise 3.4.F in

Engelking, Ryszard. "General topology." (1977)

gives further clarification and shows that 2. is not that far from a complete solution as one might think:

1. If $$X$$ is Tychonoff space (= completely regular Hausdorff space) and $$Y$$ contains a subspace homeomorphic to $$\mathbb R$$, then $$\operatorname{Map}(X,Y)$$ is compact if and only if $$Y$$ is compact and and $$X$$ is discrete.

Another useful theorem in this context is the general Ascoli Theorem as stated in Theorems 3.4.20 and 3.4.21 in Engelking's book. It gives us the following corollary:

Let $$X$$ be a $$k$$-space and $$Y$$ be a compact Hausdorff space. Then $$\operatorname{Map}(X,Y)$$ is compact if and only $$\operatorname{Map}(X,Y)$$ is evenly continuous.

Here's an open cover that has no finite subcover. Let $$U := (-1/3, 1/3)$$ and $$V := (1/6, 5/6)$$ (I'm thinking of $$S^1$$ as $$\mathbb{R}/\mathbb{Z}$$). Then $$\mathrm{Map}(S^1, S^1)$$ is covered by the open sets $$\{S([-1/k, 1/k], U) \,|\, k \in \mathbb{N}\} \cup \{S(\{0\}, V)\},$$ since a map $$S^1 \to S^1$$ must take 0 into either $$U$$ or $$V$$, and if it takes 0 into $$U$$, then being continuous it takes an interval surrounding 0 into $$U$$.

To see that this cover has no finite subcover, fix a finite collection of these open sets and let $$K$$ be the maximum value such that $$S([-1/K, 1/K], U)$$ is one of those finitely many open sets. There are continuous functions $$f : S^1 \to S^1$$ such that $$f(0) = 0 \in U$$ but $$f([-1/K, 1/K]) \nsubseteq U$$, and such functions are not contained in the union of these finitely many open sets.

Let $$\text{Cl}\left(\mathbb{Z}\right)$$ be the (discrete) space of conjugacy classes of elements of $$\pi_{1}\left(S^{1}\right)\simeq\mathbb{Z}$$ (so just $$\mathbb{Z}$$ itself...). The natural map $$\pi_{1}\ \colon\ \text{Map}\left(S^{1},S^{1}\right)\to\text{Cl}\left(\mathbb{Z}\right)$$ is continuous with non-compact image, so with non-compact domain.

(The map is continuous because the unbased loop space of $$S^{1}$$ is locally path connected. I.e., for a given unbased loop in $$S^{1}$$, there is a finite collection of compact-open mapping relationships satisfied by that unbased loop which together determine the conjugacy class in $$\pi_{1}\left(S^{1}\right)$$ induced by any unbased loop in $$S^{1}$$ that satisfies them.)

If it were we would not need Arzelà-Ascoli's theorem.

Here is an example from with a closely related space $$I=[0,1]$$. The sequence of function $$f_n(x) = x^n$$ has no convergent sub-sequence.

$$\bf{Added:}$$ We can adapt the above example to $$S^1$$. Consider the sequence of continuous functions $$f_n \colon S^1 \to S^1$$ with no convergent subsequence $$f_n(e^{2 \pi i t}) = e^{2 \pi i t^n}$$ ($$t\in [0,1]$$)