# Quotients and continuous maps

Suppose $f:\mathbb R\times \mathbb R\to \mathbb R/\mathbb Z\times \mathbb R/\mathbb Z$ where the domain is given the usual topology and the latter the quotient topology. Why then is the restriction $f|_S:S\to f(S)$, where $S=\{(x,\sqrt{5} x): x\in \mathbb R\}$ not continuous with continuous inverse?

If I am not wrong, the quotient topology is comprised of sets whose preimages are open.

I can prove that $f|_S^{-1}$ is well-defined by showing that $f|_S$ is bijective, but I don't know why they are not continuous.

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are you asking for a proof or for intuition? – Tim kinsella May 19 '12 at 10:50
the intuition is that $f|_S$ maps distant points in $S$ to nearby points in the codomain so that the $f|_S^{-1}$ would have to tear those nearby points away from each other. – Tim kinsella May 19 '12 at 10:53

Let $\alpha=\frac12\sqrt2$, and let $p_n=\left\langle n\alpha,n\alpha\sqrt5\right\rangle\in S$ for $n\in\Bbb N$; then $\{f(p_n):n\in\Bbb N\}$ is dense in $\left(\Bbb R/\Bbb Z\right)^2$. Thus, there is a strictly increasing sequence $\langle n(k):k\in\Bbb N\rangle$ in $\Bbb N$ such that $$\left\langle f(p_{n(k)}):k\in\Bbb N\right\rangle\to p_0$$ in $\left(\Bbb R/\Bbb Z\right)^2$. Let $g=f\upharpoonright S$, and let $h=g^{-1}$. If $h$ were continuous, we’d have $$\left\langle p_{n(k)}:k\in\Bbb N\right\rangle=\left\langle h\left(f(p_{n(k)})\right):k\in\Bbb N\right\rangle\to h(p_0)=\langle 0,0\rangle\;,$$ which is absurd.
The density of $\{f(p_n):n\in\Bbb N\}$ in $\left(\Bbb R/\Bbb Z\right)^2$ follows from the fact that $1,\alpha$, and $\alpha\sqrt5$ are pairwise incommensurate: none of them is a rational multiple of another.