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Consider a space of smooth real-valued functions on some domain.

For instance, have the domain be $X = \mathbb R$ or perhaps some compact domain such as $X = [0, 1]$.

For the resulting smooth function space $\text C^\infty X$, is it possible that all functions in this space are representable as a family of maps $f_t : X \rightarrow \mathbb R$, that vary continuously into each other w.r.t the parameter $t \in \mathbb R$? Intuitively this would seem to hold for smooth functions on either $\mathbb R$ or a diffeomorphism of the unit interval, but I'd like to know for sure.

If so for both of these cases, are there any notable or interesting similar situations where this property doesn't hold? Are there weaker forms of this statement in the case of failure, such as this holding for a countable subset of the original function space? Thanks in advance.

Nate
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    Are you asking about the existince of a single homotopy (not a different one for each pair of start and end functions) such that every smooth function appears as $f_t$ for some value of $t$? Or just asking about the property "any two smooth functions are homotopic to each other" (as both answerers seem to have assumed)? – Karl Dec 17 '24 at 14:44
  • @Karl The first case. I wanted the property so that I could do this. – Nate Dec 17 '24 at 16:14
  • You are basically asking if there is a continuous surjection from the unit interval to the space of all functions $C^\infty (X)$. In particular this would imply $C^\infty(X)$ is compact, which is a significant restriction. – Cheerful Parsnip Dec 17 '24 at 16:28
  • @CheerfulParsnip I was actually thinking more along the lines of $t \in \mathbb R$, even though the conventional definition of a homotopy would involve the unit interval. Perhaps I should have written that I'm looking for a continuous surjection is really from the real number line to the space of continuous functions. Truthfully, I forgot that the domain of a homotopy is the unit interval (I was thinking about homotopy geometrically as a continuous deformation of these functions into each other) – Nate Dec 17 '24 at 16:39
  • @nate No worries. I think phrasing the question as whether there is a surjection from $\mathbb R$ to $C^\infty(X)$ would probably be clearer to most people. We've already seen a couple people misread the question, not because you didn't state it clearly, but because it is an unusual way of stating things. – Cheerful Parsnip Dec 17 '24 at 17:20
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    @CheerfulParsnip I agree, I'll make the change to the title and question prompt to avoid further confusion – Nate Dec 17 '24 at 17:26

2 Answers2

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If the domain is contractible and the codomain path-connected, this is certainly true, for all functions on it are homotopic to a constant function. And if either of those things is false, then the claim is false as well, as the example consisting of a domain that's a single circle and a codomain consisting of two circles shows.

Here's a proof.

Claim: If $X$ is nonempty contractible and $Y$ is path connected, then any two (continuous) maps $f$ and $g$ from $X$ to $Y$ are homotopic.

Proof: Let $x_0 \in X$. Because $X$ is contractible, there's a homotopy $H$ from $$ i: X \to X : x \to x $$ to $$ c: X \to X : x \to x_0. $$ i.e. we have $H(x, 0) = i(x)$ and $H(x, 1) = x_0$ for all $x \in X$.

Let $K: X \times I \to Y: (x, t) \mapsto f(H(x, t))$.

Then $K(x, 0) = f(x)$ and $K(x, 1) = f(x_0)$, i.e., $K$ is a homotopy from $f$ to a constant map.

We can similarly define $L(x, t) = g(H(x, t))$ to get a homotopy from $g$ to a constant map to $g(x_0)$.

Let's name $a = f(x_0)$ and $b = g(x_0)$, and let $\gamma: I \to Y$ be a path with $\gamma(0) = a, \gamma(1) = b$, which exists because $Y$ is path-connected.

Let $M(x, t) = \gamma(t)$. That's a homotopy from "the constant map to $a$" to "the constant map to $b$".

Now consider $$ Q: X \times I \to Y : (x, t) \mapsto \begin{cases} H(x, t) & 0 \le t \le 1 \\ M(x, t-1) & 1 \le t \le 2 \\ K(x, 3-t) & 2 \le t \le 3 \end{cases} $$ and define $R(x, t) = Q(x, 3t)$. Then $R$ is a homotopy from $f$ to $g$. QED

John Hughes
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    I thought the OP was looking for a single homotopy (not a different one for each pair of start and end functions) such that every smooth function appears at some value of $t$. We could try concatenating many homotopies of the kind constructed here, but it seems to me there could be too many smooth functions for that to work. – Karl Dec 17 '24 at 14:40
  • I thought that was a possibility as well, but apparently my answer was what OP was looking for. I agree that the "too many functions" argument might work for some pair of spaces, where the cardinality of $Maps(A, B)$ is larger than the cardinality of the interval. – John Hughes Dec 17 '24 at 14:57
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    @JohnHughes I was looking for the first case. I apologize for errantly marking what you wrote as the answer. I saw "yes" and didn't see that you were arguing for the "any two functions are homotopic case". If there's a failure, are we at least guaranteed a countable (hopefully dense) set of functions that are all part of one homotopy? – Nate Dec 17 '24 at 16:18
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    Well...we can find a homotopy that passes through uncountably many constant functions...indeed, $L$ already does this. If $\gamma$ were a space-filling curve, we could even make $L$ pass through all the constant functions. Sadly, that's only a tiny fraction of the set of all continuous functions. And, as this https://math.stackexchange.com/questions/2915880/is-the-cardinality-of-mathcalc0-1-the-same-as-the-cardinality-of-mathbb answer points out, if $X$ is $I$ (or probably any compact set) the set $C^0(X)$ has the same cardinality as the reals. But is there a continuous map from $I$ ... – John Hughes Dec 17 '24 at 20:24
  • ..onto it? I sure doubt it. Maybe someone good at analysis can answer this question for the case $X = Y = I$ at least. – John Hughes Dec 17 '24 at 20:24
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Since you have used the tag homotopy-theory, you might be interested to know that the entire topic of homotopy-theory can be thought of as the study of examples and counterexamples to a generalized version of your property, where the codomain $\mathbb R$ in your question is replaced by more general spaces. So when your title question is considered in that generalized manner, a complete resolution is entirely unreasonable, but a special case might be instructive.

For example, let's consider smooth functions from $X$ to $S^1$ where $X$ is a smooth manifold. It is a theorem that the set of homotopy classes of smooth functions from $X$ to $S^1$ is in one-to-one correspondence with the first cohomology group $H^1(X,\mathbb Z)$ with $\mathbb Z$ coefficients. In other words, the property that you ask about --- that any two smooth functions $X \to S^1$ may be continuously deformed to one another --- is mathematically equivalent to saying that $H^1(X,\mathbb Z)$ is the trivial group.

So you get many examples of your property using manifolds for which $H^1(X,\mathbb Z)$ is trivial: spheres $S^n$ with $n \ge 2$; real projective spaces $\mathbb RP^n$ with $n \ge 2$; complex projective spaces $\mathbb CP^n$, ...

And you get many counterexamples using using manifolds for which $H^1(X,\mathbb Z)$ is nontrivial: all closed surfaces of non-positive Euler characteristic (for example the Klein bottle); all toruses $\mathbb T^n = (S^1)^n$ of dimensions $n \ge 1$; ...

Lee Mosher
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