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For topological spaces $X,Y$ let $\text{Cont}(X,Y)$ be the collection of continuous functions $f:X\to Y.$ We endow $\text{Cont}(X,Y)$ with the topology inherited from the product topology on $Y^X.$

Are there spaces $X,Y$ such that $X$ has more than one point and $Y\not\cong\mathbb{R}$ such that $\mathbb{R}\cong\text{Cont}(X,Y)$?

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I must admit that I hesitated answering this question, but here it is.

The answer is "no". Assume there exist topological spaces $X$ and $Y$ such that $C(X,Y)\simeq \mathbb{R}$ and $Y\not\simeq\mathbb{R}$. Identifying $C(X,Y)$ with $\mathbb{R}$ and $Y$ with the constant functions in $Y^X$ we consider $Y$ as a closed subsapce of $\mathbb{R}$. Considering the surjection $\mathbb{R} \to C(X,Y)\to C(\{x\},Y) \to Y$, we see that $Y$ is connected. Thus $Y\subset\mathbb{R}$ is a closed convex subset. As $Y\not\simeq \mathbb{R}$ there must exist an extreme point $y\in Y$. Note that $C(X,Y)$ is a convex subset of $C(X,\mathbb{R})$ and the constant function $y$ is an extreme point of it. It follows that $C(X,Y)-\{y\}$ is also convex, hence contractible. But $C(X,Y)-\{y\}$ is homeomorphic to $\mathbb{R}$ minus a point. This is a contradiction.

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    $\begingroup$ Sorry to be dense, but I must admit that I'm not following the final sentence. First, I don't know how to parse postcomposing an object $C(X, Y)$ with a contraction map. Second, I'm not seeing where $\mathbb{R} - \{1\}$ is coming from. Third, we didn't use anything about $X$ having more than one point that I can see (maybe we don't actually need that condition?). $\endgroup$ – Todd Trimble Jan 8 '18 at 14:36
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    $\begingroup$ In fact, that X has more than one point follows from the assumption Y is not homeomorphic to $\mathbb{R}=C(X,Y)$ $\endgroup$ – Pietro Majer Jan 8 '18 at 15:06
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    $\begingroup$ @Todd, Third thing first, we didn't use anything about $X$ because we didn't need to. Second, $\mathbb{R}-\{1\}$ stands for $C(X,Y)$ minus the constant function 1, and for your first complain: that was my sloppy way to explain the obvious fact that $C(X,Y)-\{y\}$ is contactible where $Y$ is a convex set, $y\in Y$ is an extreme point and $y\in C(X,Y)$ represents the constant function. That seems obvious enough to me, so I allowed myself some informal presentation... $\endgroup$ – Uri Bader Jan 8 '18 at 16:44
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    $\begingroup$ Indeed, the proof of Uri Bader was correct (exploiting a very nice observation with the extreme point of $C(X,Y)$). Initially I did not catch the idea and because of that wrote my proof (actually a continuation of the proof of Uri Bader). $\endgroup$ – Taras Banakh Jan 8 '18 at 17:52
  • $\begingroup$ Uri, thanks for the edit. I wasn't trying to nitpick -- I honestly did find the original hard to follow. Nice proof. $\endgroup$ – Todd Trimble Jan 8 '18 at 17:55
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Following the answer of Uri Bader, we can show that $Y$ is a retract of the real line, so can be identified with a closed convex subset of $\mathbb R$. Without loss of generality we can assume that $0,1\in Y$ and hence $[0,1]\subset Y$. It follows that the function $C(X,Y)$ is a convex subset of $Y^X\subset \mathbb R^X$ and $C(X,[0,1])\subset C(X,Y)$. The assumption $Y\not\cong\mathbb R\cong C(X,Y)$ implies that $C(X,Y)$ contains a non-constant function $f$. Consider the constant functions $\mathbf 0:X\to\{0\}\subset Y$ and $\mathbf 1:X\to\{1\}\subset Y$ and observe that the set $T=\{\mathbf 0,\mathbf 1,f\}\subset C(X,Y)\subset Y^X\subset\mathbb R^X$ is affinely independent and its convex hull $conv(T)\subset C(X,Y)$ is homeomorphic to the 2-dimensional symplex, which cannot be contained in the real line $\mathbb R\cong C(X,Y)$. This contradiction completes the proof.

The negative answer can also be deduced from the following theorem. We recall that a topological space $X$ is functionally Hausdorff if for any distinct points $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ such that $f(x)\ne f(y)$.

Theorem. If for non-empty topological spaces $X,Y$ the function space $C(X,Y)$ is functionally Hausdorff and path-connected, then either $C(X,Y)$ is homeomorphic to $Y^n$ for some $n\in\mathbb N$ or $C(X,Y)$ contains a topological copy of the Hilbert cube.

Proof. The space $Y\cong C(\{x\},Y)$ is functionally Hausdorff and path-connected, being a retract of the functionally Hausdorff path-connected space $C(X,Y)$. If $Y$ is a singleton, then $C(X,Y)\cong Y^1$ is a singleton, too. So, we assume that $Y$ contains more than one point. In this case $Y$ contains a subspace $I$, homeomorphic to the closed interval $[0,1]$.

Consider the canonical map $\delta:X\to Y^{C(X,Y)}$, $\delta:x\mapsto (f(x))_{f\in C(X,Y)}$. If the image $\delta(X)$ is finite of cardinality $n$, then $C(X,Y)$ is homeomorphic to $Y^n$ (since each function $f\in C(X,Y)$ is constant on each set $\delta^{-1}(y)$, $y\in \delta(X)$).

So, we assume that the set $\delta(X)$ is infinite. Taking into account that the space $Y^{C(X,Y)}$ is functionally Hausdorff, we can construct a continuous map $g:Y^{C(x,Y)}\to I$ such that the image $Z=g(\delta(X))$ is infinite. The surjective continuous map $p:=g\circ\delta:X\to Z$ induces a continuous injective map $p^*:C(Z,I)\to C(X,I)$, $p^*:f\mapsto f\circ p$. It is easy to see that the function space $C(Z,I)\subset I^Z$ contains a topological copy $Q$ of the Hilbert cube $I^\omega$. Then $p^*(Q)$ is a topological copy of the Hilbert cube in $C(X,I)\subset C(X,Y)$.

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  • $\begingroup$ This proof I follow; +1. $\endgroup$ – Todd Trimble Jan 8 '18 at 15:04
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    $\begingroup$ Thanks Taras for this nice presentation! I accepted Uri's answer because it was the original one and I understood it after the edit $\endgroup$ – Dominic van der Zypen Jan 8 '18 at 19:56
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    $\begingroup$ @DominicvanderZypen No problems. You made a correct decision. Uri Bader was the first and had a brilliant idea. $\endgroup$ – Taras Banakh Jan 8 '18 at 20:01

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