Is the preimage of $f(x) = x_1 x_2 \dots x_n$ connected?

Question

Suppose $x = (x_1, \dots, x_n) \in \mathbb R^n$ and we define $f : \mathbb R^n \to \mathbb R$ by $(x_1, \dots, x_n) \mapsto x_1 x_2 \cdots x_n$. If $E \subset \mathbb R$ is connected, is $f^{-1}(E)$ connected? I tried a few examples and it seems so.

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I used a probably bad notation initially by $x = (x^1, \dots, x^n)$. I don't why I used that. Probably because one of my textbooks used this notation but I didn't consider situations. Also I meant to ask when $E$ is not a singleton.

Answer 1

For $n=1$, the map is just the identity so there is nothing to show. For $n>1$, the preimage of a connected set $\emptyset \neq E\subset \mathbb R$ under the map $f\colon \mathbb R^n\to \mathbb R, (x_1,\dots,x_n)\mapsto \prod_{i=1}^nx_i$ is connected if and only if $0\in E$. Let $A$ denote this preimage.

Assume $0\notin E$, then for any $e\in E$, the two points $(e,1,\dots,1)$ and $(-e,-1,1,\dots,1)$ lie in two different components $A$, because they are separated by the hyperplane $\{x_1=0\}$ which is disjoint from $A$ (because it is mapped to $0\notin E$).

Conversely, if $0\in E$, given any point $x=(x_1,\dots,x_n)$ we can consider the path $[0,1]\to \mathbb R^n, t\mapsto tx$ which connects $0$ to $x$. For every $t\in [0,1]$ we have that $f(tx)=t^nf(x)$ which is contained in the interval $[0,f(x)]$ (or $[f(x),0]$, if $f(x)$ is negative) which is contained in $E$ because $0$ and $f(x)$ are (and $E$ is connected). Hence the path connects $0$ and $x$ in the preimage $A$, hence $A$ is connected.

Answer 2

Not for all $n$, $E$. To have a clear counterexample, we make a choice of an $n>1$ and of an $E$ that

• does not contain the zero as an element,
• is simple enough to finish in few rows.

For $E=(0,\infty)$ and $n>1$ the preimage of $E$ is the set of all $x\in \Bbb R^n$ which have an even number of negative components, and no zero component. This set is not connected, because its image through the continuous map $\pi_1$ = projection onto the first component "$x^1$" is $\Bbb R-\{0\}$, $$ \pi_1(\ f^{-1}(E)\ )\ =\ \Bbb R-\{0\}\ , $$ a set which is not connected.

(But a continuous function maps connected sets to connected images.)

If $E$ contains the zero element... but this is an other question.