# Is a space of functions between two simply connected spaces connected itself?

EDITED:

As some commenters have suggested, this question may be viewed as "homotopy in disguise". Perhaps it would be best to give the underlying spaces more structure.

Let $X$ and $Y$ be simply connected locally compact Hausdorff spaces (like, e.g., finite-dimensional Banach/Hilbert spaces or $\mathbb R^{n>2}$). Let $Y^X$ denote the space of all morphisms $f:X\longrightarrow Y$ with the compact-open topology. Obviously when $X,Y$ are Euclidean, the morphisms are continuous functions, when they are topological vector spaces, they are continuous linear maps, etc.

The fundamental group of $X,Y$ is trivial, and the exponential object exists.

The question: Is $Y^X$ a connected space?

• – Joshua Ruiter Apr 19 '17 at 13:17
• Thanks, I stumbled upon this before posting, but I can't find a way to conclude that $C(X,Y)$ (or $Y^X$) is in fact connected under the compact-open topology. – Dawid K Apr 19 '17 at 13:22
• If you focus instead on path-connected components then philosophically speaking, a path $\gamma \colon [0,1] \rightarrow C(X,Y)$ should be the same as a homotopy between the two functions $\gamma(0), \gamma(1) \colon X \rightarrow Y$ so path-connected components should be classified by homotopy classes of maps between $X$ and $Y$. This indeed holds under reasonable assumptions on $X,Y$ which hold for $X = Y = \mathbb{R}$ and since $\mathbb{R}$ is convex, $C(\mathbb{R},\mathbb{R})$ is path-connected under the compact open topology and also connected. I'm not sure about the other case nor – levap Apr 19 '17 at 14:04
• whether there might be a difference between path-connected components and connected components for function spaces but this might give you some keywords to search for. – levap Apr 19 '17 at 14:05
• Homotopy is way over my head, but I'll try to look into this. I was also thinking about linking this somehow to the product space topology of $X\times Y$, since it is known that the product space is connected when the underlying sets are connected, and moreover functions can be represented as Cartesian products. Not sure how to square it, though. – Dawid K Apr 19 '17 at 14:23

No, this is very false in general. For example, if $X=Y=S^n$ ($n>0$), the connected components of the space $Y^X$ of continuous maps from $X$ to $Y$ are in bijection with $\mathbb{Z}$ (the integer corresponding to a map $f:X\to Y$ is known as the degree of $f$). For $n>1$, $X$ and $Y$ are simply connected. In general, determining the connected components such spaces $Y^X$ (when $Y$ and $X$ are reasonably nice, at least) is a deep geometric problem and is one of the central problems of the entire field of algebraic topology. Just as an example, classifying all of the connected components of $Y^X$ in the case $Y=S^m$ and $X=S^n$ for arbitrary values of $m$ and $n$ is fantastically difficult and the answer is so complex we will probably never have any satisfactory complete description of it (see https://en.wikipedia.org/wiki/Homotopy_groups_of_spheres for an overview of the problem).

One case where it is true is if $X$ or $Y$ is contractible: we say $X$ is contractible if there exists a continuous map $H:X\times [0,1]\to X$ such that $H(x,0)=x$ for all $x$ and $H(x,1)$ is constant (i.e., the same for all $x$). For example, if $X=\mathbb{R}^n$, then $X$ is contractible via the map $H(x,t)=(1-t)x$.

If $X$ is contractible, then for any $f\in Y^X$, there is a continuous map $F:[0,1]\to Y^X$ given by $F(t)(x)=f(H(x,t))$. Note that $F(0)=f$, and $F(1)$ is the constant function with value $f(H(-,0))$. It follows that $Y^X$ is path-connected, since every element of $Y^X$ can be connected to a constant function by a path, and all constant functions can be connected by paths since you have assumed $Y$ is simply connected (in particular, path-connected). A similar argument shows that when $Y$ is contractible, $Y^X$ is path-connected (in fact, more strongly, $Y^X$ is contractible, without needing any hypothesis like path-connectedness on $X$).

A version of the argument above also works if you are only considering continuous linear maps between topological vector spaces, since the contraction $H(x,t)=(1-t)x$ only passes through linear maps.

• I will wait just a little before accepting, but this is a fantastic answer! Just a couple of questions. The map $F$ is 1-ary, but you write $F(t)(x)$ - does it mean that $x$ is fixed (constant), e.g. $F_x(t)$? Also you mention that for TVect the argument needs another version – can you elaborate just a bit? I was thinking about using a simple argument when reasoning in the category of TVect: since $Y^X$ is an internal hom, and so an object of the category, it must also be a topological vector space (e.g., by tensor product), and thus connected by definition. – Dawid K Apr 20 '17 at 19:22
• Remember that the codomain of $F$ is $Y^X$, so $F(t)$ is an element of $Y^X$: that is, a function $X\to Y$. So you can evaluate $F(t)$ at an element $x\in X$: that's what $F(t)(x)$ is. You're right that in the vector space case you can also just say that $Y^X$ is a topological vector space and hence connected. – Eric Wofsey Apr 20 '17 at 21:26

If you ask about path connectness instead of connectness then this is a "homotopy in disguise" (side question: are path-connectness and connectness equivalent in $Y^X$?). Indeed, for reasonably good spaces (i.e. $X$ locally compact Hausdorff) every path

$$\lambda:[0,1]\to Y^X$$

induces a homotopy

$$H:[0,1]\times X\to Y$$ $$H(t, x)=\lambda(t)(x)$$

(note that $\lambda(t):X\to Y$ is a function so $\lambda(t)(x)$ makes sense) and vice versa: every homotopy $$H:[0,1]\times X\to Y$$

induces a path

$$\lambda:[0,1]\to Y^X$$ $$\lambda(t)=H_t$$

where $H_t(x)=H(t,x)$.

Under the assumption that $X$ is locally compact Hausdorff this correspondence actually defines a homeomorphism $(Y^X)^{[0,1]}\to Y^{X\times [0,1]}$ .

So the question

Is $Y^X$ a path-connected space?

Are all continous maps $X\to Y$ homotopic?
So indeed, this does not have to hold even when both $X,Y$ are simply connected (e.g. $X=Y=S^2$). Although it does hold if for example $Y$ is contractible.