# Submanifolds of a space of functions?

Let $f:\mathbb{R}\rightarrow(\mathbb{R}\rightarrow\mathbb{R})$ be a function mapping a real number uniquely into the set $\mathbb{F}$ of total functions from $\mathbb{R}$ to $\mathbb{R}$. $\mathbb{F}$ is a topological space with the following definition of a neighborhood topology $\mathbf{N}_\mathbf{y}:\mathbb{F}\rightarrow2^\mathbb{F}$ for a fixed tuple $\mathbf{y}\in \mathbb{R}^n$:

$$\mathbf{N}_\mathbf{y}(g)=\{\{g'\mid g'\in\mathbb{F},\;\sum^n_{i=1}|g(\mathbf{y}_i)-g'(\mathbf{y}_i)|\leq r\}\mid r\in\mathbb{R}\}$$ (ie. for every $r \in \mathbb{R}$ the set of functions whose accumulated abs. difference to $g$ sampled at $n$ fixed points is not greater than $r$)

Then, can I say that the range of $f$ is a 1-dimensional submanifold of $\mathbb{F}$?

Is it if I add the constraint that $f(x)(y)$ is continuous in $x$ for every fixed $y\in\mathbb{R}$? My intuition is that then it is a submanifold because $f$ is invertible (wrt. its range) and because it is topology-preserving due to $f(x)(y)$ being continuous in $x$.

But honestly I'm not sure if that's enough or even correct; I don't know a lot about topology (yet).

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Before mentioning submanifolds, you should first be more precise about the space in which it lies. So what's the topology on the space of maps $\mathbf{R^n}\to\mathbf{R}$? Do you really want all maps or just continuous ones? – ZulfiqarIII Jan 24 '13 at 11:54
Just to illustrate what @ZulfiqarIII said: if you choose the discrete topology on the set of functions, it will not admit any subspace homeomorphic to $\mathbb{R}$ (continuous function from $\mathbb{R}$ to any discrete space is single-valued). – Willie Wong Jan 24 '13 at 12:08
Thanks for the comments; they prompted me to think about the question more deeply. I've simplified and added a definition of topology for functions from $\mathbf{R}$ to $\mathbf{R}$. – Johannes Bauer Jan 24 '13 at 13:48
"because $f$ is reversible"... do you mean invertible? and why is this true? – user53153 Jan 24 '13 at 19:47
Oops. Yes, invertible. I'm correcting that. It is (wrt. its range), because I said it maps a real uniquely into $\mathbb{F}$. I'll also make that `wrt. its range' part clearer although I don't know if that's correct terminology. – Johannes Bauer Jan 25 '13 at 7:45

If you take your definition of $\mathbb{F}$, with the neighborhoods $\mathbf{N}_y$, then $\mathbb{F}$ is not Hausdorff, and you can get counterexamples pretty easily.
Let $z\in\mathbb{R}$ such that $z$ is not one of the $y_i$. Let $\phi$ be a bump function on $\mathbb{R}$ such that $\phi(z) = 1$ and $\phi(y_i) = 0$ for every $y_i$.
Let $f(x)(y) = x\phi(y)$. Since $f(x)(z) = x$ we see that $f$ is injective.
On the other hand, since $f(x)(y_i) = 0$ for every $y_i$ we have that the induced topology on $f(\mathbf{R}) \subset \mathbb{F}$ is the indiscrete topology. Hence $f(\mathbf{R})$ can not be locally or globally homeomorphic to $\mathbf{R}$, and it is not a manifold (and hence not a submanifold).
Thanks. Still digest that and trying to see what it means for my case (using self-organizing maps in order to learn the mapping from manifolds to $\mathbf{R}^n$). – Johannes Bauer Jan 25 '13 at 11:52