# No surjection onto function space for manifold X

[*Note: highly technical. Skip if topology is not your thing*]

In his post on [formal open problems in decision theory](https://www.lesswrong.com/posts/5bd75cc58225bf06703753a9/formal-open-problem-in-decision-theory), Scott asked whether there could exist a topological space $X$ and a continuous surjection $s$ from $X$ to $C(X,I)$. Here, $I$ is the closed unit interval $[0,1]$ and $C(X,I)$ is the set of continuous functions from $X$ to $I$.

I thought I had [an argument](https://www.lesswrong.com/posts/5bd75cc58225bf06703753a9/formal-open-problem-in-decision-theory#pXcgwa4L7SeCESY9R) for how $\mathbb{R}-\mathbb{N}$ could be such an $X$. But that argument is wrong, as I'll demonstrate in this post. Instead I will show that:

* Let $X$ be a manifold (with or without boundary), or a union of finitely or countably many manifolds. Then there is no continuous surjective map from $X$ to $C(X,I)$.

By "union", imagine the manifolds lying inside Euclidean space (or, more generically, inside a [metric space](https://en.wikipedia.org/wiki/Metric_space)), not necessarily disjoint, and taking their unions there. Note that there are many examples of such $X$s - for example, the rationals within the reals (being countable unions of points, which are trivial manifolds).

In fact, I will show the more general:

* If $X$ is [Fréchet–Urysohn](https://en.wikipedia.org/wiki/Sequential_space#Fr%C3%A9chet-Urysohn_space), and [$\sigma$-compact](https://en.wikipedia.org/wiki/%CE%A3-compact_space), then there is no continuous surjective map from $X$ to $C(X,I)$.

To see that this more general result implies the one above, note that [manifolds are $\sigma$-compact](https://math.stackexchange.com/questions/1200963/locally-euclidean-hausdorff-topological-space-is-topological-manifold-iff-sigm), and that if $X_i$ is $\sigma$-compact, it can be covered by countably many compact sets, so $\bigcup_{i\in\mathbb{N}} X_i$ can be covered by countably many sets of countably many compact sets, which is just countably many. Finally, [all metric spaces are Fréchet–Urysohn](https://arxiv.org/pdf/math/0412558.pdf).

Fréchet–Urysohn basically means "convergence of subsequences makes sense in the topology", and is not a strong restriction; indeed all [first-countable](https://en.wikipedia.org/wiki/First-countable_space) spaces are Fréchet–Urysohn.

# Proof

## A note on topologies

Now $C(X,I)$ is well-defined as a set, but it needs a topology to discuss issues of continuity. There are three natural topologies on it: the topology of [uniform convergence](https://en.wikipedia.org/wiki/Uniform_convergence), the [compact-open topology](https://en.wikipedia.org/wiki/Compact-open_topology) (which, on this set, is equal to the topology of [compact convergence](https://en.wikipedia.org/wiki/Compact_convergence)), and the topology of [pointwise convergence](https://en.wikipedia.org/wiki/Pointwise_convergence).

The one most people use, and that I'll be using, is the compact-open topology.

These are refinements of each other as so:

* uniform convergence $\supset$ open-compact $\supset$ pointwise convergence.

Thus, if we could find a surjective $s:X \to C(X,I)$ in the uniform convergence topology, it would still be continuous in the compact-open topology. Conversely, if we could show that no such maps exist in the pointwise convergence topology, there would also be no maps in the compact-open one. Unfortunately, the partial results I've found are opposite: I believe I can show that no maps exist for general $T_4$ $X$'s for uniform convergence, and I have a [vague argument](https://www.lesswrong.com/posts/5bd75cc58225bf06703753a9/formal-open-problem-in-decision-theory#5bd75cc58225bf06703753b1) that may allow us to construct one in the pointwise convergence topology. Neither of these help here.

Anyway, onwards and upwards!

## The steps of the proof

The proof will have lots of technical terminology (with links) but will be short on detailed explanations of this terminology.

It will first assume that $X$ is Fréchet–Urysohn, $\sigma$-compact, and a [$T_4$](https://en.wikipedia.org/wiki/Normal_space) space. Once the proof is completed, it will then go back and remove the $T_4$ condition.

Why do it in that order? Because without the $T_4$ condition, it is a proof by contradiction, assuming a continuous surjection $s:X \to C(X,I)$ to do most of the work. I have nothing against proofs by contradiction, but, if ever anyone wants to refine or extend my proof, I want to make clear which results are really true for $X$, and which ones only arise via the contradiction.

The proof will proceed by these steps:

* Excluding discrete spaces, and finding compact $Y\subset X$.
* The restriction map $C(X,I) \to C(Y,I)$ is continuous surjective.
* Compact subsets of $C(Y,I)$ have empty interior.
* $C(Y,I)$ is a Baire space.
* There is no surjective continuous map $s: X \to C(X,I)$.
* Removing the $T_4$ condition.

## Excluding discrete spaces, and finding compact $Y\subset X$

If $X$ is discrete, then $C(X,I) = I^X > 2^X$, which has strictly higher cardinality than $X$, so surjective maps $X \to C(X,I)$ are impossible at the set level.

So put the discrete sets aside. Since $X$ is not exclusively made up of [isolated points](https://en.wikipedia.org/wiki/Isolated_point), it must contain a [limit point](https://en.wikipedia.org/wiki/Limit_point); call it $y_\infty$. Since $X$ is Fréchet–Urysohn, there exists a sequence $\{y_i\}_{i\in\mathbb{N}}$, $y_i \neq y_\infty$, that converges on $y_\infty$.

Then define $Y = \{y_i\}_{i\in\mathbb{N}} \cup \{y_\infty\}$; it's clear that it is compact in the subspace topology. Since $X$ is also [$T_2$](https://en.wikipedia.org/wiki/Hausdorff_space), $Y$ [must be closed](http://mathonline.wikidot.com/compact-sets-in-hausdorff-topological-spaces) in $X$.

## The restriction map $C(X,I) \to C(Y,I)$ is continuous surjective

Any function $f:X \to I$ is a map $f:Y \to I$ by restriction, so there is a map $r: C(X,I) \to C(Y,I)$.

Since $X$ is $T_4$ and $Y$ is closed, for any continuous function $f: Y \to I$, there exists, by the [Tietze extension theorem](https://en.wikipedia.org/wiki/Tietze_extension_theorem), a continuous $F: X \to I$ such that $f(y) = F(y)$ for all $y \in Y$. So $r$ is surjective.

We now want to show that $r$ is also continuous.

A subbase for $C(Y,I)$ [consists](https://en.wikipedia.org/wiki/Compact-open_topology#Definition) of all $V_{Y}(K,U)\subset C(Y,I)$ where $K\subset Y$ is compact in $Y$ and $U\subset I$ is open, and $f\in V_{Y}(K,U)$ iff $f(K)\subset U$.

Note that since $Y$ has the subspace topology in $X$, these $K$ must be compact in $X$ as well. So define $V_X(K,U)$, consisting of continuous functions $F$ from $X$ to $I$ with $F(K)\subset U$. Then $p^{-1}(V_{Y}(K,U))=V_X(K,U)$, and thus the pre-image of the sets of this subbasis is open.

Since it [suffices to check](https://en.wikipedia.org/wiki/Subbase#Results_using_subbases) continuity on a subbasis, this shows that $r$ itself is continuous.


## Compact subsets of $C(Y,I)$ have empty interior

We aim to show, using [Ascoli's Theorem](https://people.math.osu.edu/carr.520/files/11_Compact-Open_and_Ascoli.pdf), that any compact subset of $C(Y,I)$ has empty [interior](https://en.wikipedia.org/wiki/Interior_(topology)) - thus contains no open sets.

The subbasis for the topology on $Y$ consists of sets of the form $V(K,U)$, where $K$ is compact in $Y$, $U$ is open in $I$, and $f \in V(K,U)$ iff $f(K) \subset U$.

The compact subsets of $Y$ are a) finite collections of points $y_i$, $i<\infty$, b) collections of points that include all points $y_i$ for $i>N$, along with $y_\infty$, and c) $y_\infty$ itself.

Let $W$ be an intersection of finitely many $V(K,U)$. If $W$ is non-empy, it is defined by an $N \in \mathbb{N}$ and a collection of sets $\{U_i\}_{i<N} \cup \{U_{<\infty}\} \cup \{U_\infty\}$ open in $I$. Then $f$ belongs to $W$ iff:

* $\forall i < N: f(y_i) \in U_i$,
* $\forall i \geq N: f(y_i) \in U_{<\infty}$,
* $f(y_\infty) \in U_\infty$.

Since $W$ is non-empty and $f$ is continuous, we must have $\overline{U_{<\infty}}$ and $U_\infty$ having non-zero intersection. This actually implies that $U_{<\infty}$ and $U_\infty$ have non-zero intersection.

Note that this includes all possible $W$, as we can always set $U_i = I$ (or $U_{<\infty}=I$ or $U_{\infty}=I$), thus removing that particular restriction.

We want to show that such an $W$ cannot be [equicontinuous](https://en.wikipedia.org/wiki/Equicontinuity) at $y_\infty$.

To do so, fix points $u_i \in U_i$ for $i< N$, and $u_\infty, u' \in U_{<\infty} \cap U_\infty$, with $u_\infty \neq u'$

Then consider the functions $f_m \in W$, for $m>N$, defined by:

* $f_m(y_i) = u_i$ for $i < N$,
* $f(y_i) = u'$ for $N \leq i \leq m$,
* $f(y_i) = u_\infty$ for $i>m$.

These $f_m$ are all in $W$, and continuous (since $Y$ is discrete except at $y_\infty$). If $W$ were equicontinuous then for all $\epsilon > 0$, there would exists an open set $U$ in $Y$ containing $y_\infty$ such that $|f_m(y) - f_m(y_\infty)| < \epsilon$ for all $m$ and all $y \in U$.

Since $U$ is open, there exists a $y_n \in U$, $n \neq \infty$. Set $\epsilon = |u' - u_\infty|/2$; then $|f_n(y_n) - f_n(y_\infty)| = | u' - u_\infty| > \epsilon$.

Since $W$ is not equicontinuous, and $Y$ is both compact and [Hausdorff](https://en.wikipedia.org/wiki/Hausdorff_space) (and hence pre-compact Hausdorff), $W$ cannot be contained in a compact subset of $C(Y,I)$, by Ascoli's Theorem.

Any open set in $C(Y,I)$ consists of unions of sets of the type $W$, so no (non-empty) open set is contained in compact subset of $C(Y,I)$.

## $C(Y,I)$ is a Baire space

By the [Baire category theorem](https://en.wikipedia.org/wiki/Baire_category_theorem), if $C(Y,I)$ is a [complete metric space](https://en.wikipedia.org/wiki/Complete_metric_space), then it is a [Baire space](https://en.wikipedia.org/wiki/Baire_space).

Since $Y$ is compact, the compact-open and the [uniform convergence](https://en.wikipedia.org/wiki/Uniform_convergence) topology [match up](https://en.wikipedia.org/wiki/Compact-open_topology#Properties) on $C(Y,I)$. So $C(Y,I)$ has a uniform norm $d$ compatible with its topology: $d(f,g) = \sup_{y \in Y} |f(y)-g(y)|$.

So $C(Y,I)$ is thus a metric space. Since $I$ is complete, the [uniform limit theorem](https://en.wikipedia.org/wiki/Uniform_limit_theorem) shows $C(Y,I)$ is a complete metric space (and hence a Baire space).

## There is no surjective continuous map $s: X \to C(X,I)$

Let $s: X \to C(X,I)$ be any continuous map.

Since $X$ is $\sigma$-compact, $X = \bigcup_{i\in\mathbb{N}} K_i$ for compact $U_i$. The set $r \circ s(K_i)$ is compact in $C(Y,I)$, since the continuous image of compact spaces are compact.

Let $L_i = C(Y,I) - r \circ s(K_i)$. Since compact sets in $C(Y,I)$ must have empty interior, the complement of $L_i$ has empty interior. [Therefore](https://en.wikipedia.org/wiki/Interior_(topology)#Interior_operator) the $L_i$ are dense in $C(Y,I)$.

The space $r \circ s (X)$ is equal to: 

* $\cup_{i\in\mathbb{N}} (r\circ s (K_i)) = \cup_{i\in\mathbb{N}} (L_i)^c = \left(\cap_{i\in\mathbb{N}} L_i\right)^c$, where $c$ denotes taking the complement in $C(Y,I)$.

But, since $C(Y,I)$ is a Baire space, $\cap_{i\in\mathbb{N}} L_i$ is dense, hence certainly not zero, and so $r \circ s (X) \neq C(Y,I)$.

Since $r$ is surjective $C(X,I) \to C(Y,I)$, the $s$ cannot be surjective onto $C(X,I)$, proving the result.

# Removing the $T_4$ condition

We only used $T_4$ to prove the properties of the subsequence $Y$. Without $T_4$, we can use the continuous surjection $s: X \to C(X,I)$ itself. So now assume that such an $s$ exists, and we shall try and find a contradiction.

The $\sigma$-compact property implies $X$ is [Lindelöf](https://en.wikipedia.org/wiki/Lindel%C3%B6f_space). Lindelöf is preserved by continuous maps, so because of $s$, $C(X,I)$ is also Lindelöf.

Because $I$ is [$T_{3}$](https://en.wikipedia.org/wiki/Regular_space), then [so is](https://en.wikipedia.org/wiki/Compact-open_topology#Properties) $C(X,I)$. But any Lindelöf regular space is normal (see [theorem 14](https://math631spring2011.files.wordpress.com/2011/01/axiomsofseparation3.pdf)), hence, since $C(X,I)$ is also [$T_2$](https://en.wikipedia.org/wiki/Hausdorff_space), then it must be Hausdorff normal, ie $T_4$.

Now $C(X,I)$ contains at least the constant functions, so must be of uncountable cardinality. Since $X = \bigcup_{i\in\mathbb{N}} K_i$ is a countable union of spaces, there exists a $K_i$ such that $s(K_i)$ contains a set $W$ with infinitely many points. Pull these infinitely many points back to $K_i$, using the [axiom of choice](https://en.wikipedia.org/wiki/Axiom_of_choice) to choose a set $V\subset K_i$ such that for each $w\in W$, there is a unique $v \in V$ with $s(v)=w$.

Now, $K_i$ is compact, which means that it is countably compact. Now, [Fréchet–Urysohn is hereditary](https://arxiv.org/pdf/math/0412558.pdf), meaning that it applies to subspaces as well, so $K_i$ is Fréchet–Urysohn, which means that it is sequential. For sequential spaces, countably compact implies [sequentially compact](https://en.wikipedia.org/wiki/Sequentially_compact_space) (see [theorem 10](http://www.cs.cmu.edu/~yaoliang/mynotes/compact.pdf)).

Thus we can pick a sequence in $V$, and, passing to a subsequence if necessary, we can find a sequence $\{y_i\}_{i\in\mathbb{N}}$ converging to a point $y_\infty$ that is not equal to any of the $y_i$. Define $Y$ as before.

The set $s(Y)$ must also be a convergent subsequence.

Let $f \in C(Y,I)$, then $f \circ s^{-1} \in C(s(Y),I)$ (note that $s^{-1}$ is well defined here, as it is a bijection between $Y$ and its image). Because $C(X,I)$ is itself $T_4$, there exists a function $F \in C(C(X,I),I)$ that is equal to $f \circ s^{-1}$ on $s(Y)$. Then the function $F \circ s$ is equal to $f$ on $Y$, and is an element of $C(X,I)$. Thus the restriction map $r:C(X,I) \to C(Y,I)$ is still surjective, and the topology on $Y$ (indeed on any sequence tending to a point) is the same as before. The argument for $r$ being continuous goes through as before.

Then the rest of the proof proceeds as above.