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I have to say that I'm quite lost on this one. Let $c(t) = \left(\frac{t}{1 + t^4}, \frac{t}{1 + t^2}\right)$ be the regular injective path. In order to finish the proof on why $c(t)$ is no homeomorphic to its image set, my source claims that "[...] it is enough to observe that $c(t) \to (0, 0)$ as $t \to \infty$". Why is it "enough" to observe this? Why does this show that $c(t)$ is not continuous?

Epsilon Away
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    So, $c(-\infty) = c(0) = c(\infty)$. Consider the diagram here for how having the two ends approaching a point in the middle causes problems for homeomorphicity. In fact, it is enough to see that either end is abutting the middle of the image curve. – Eric Towers Nov 05 '21 at 18:57
  • @EricTowers While I don't quite understand why "it is enough to see that either end is abutting the middle of the image curve", is the reason why the inverse of $c$ is not continuous due to the fact that while $\mathbb{R}$ is open, and under a continuous mapping any open set in the range has an open pre-image in the domain, the pre-image of $\mathbb{R}$ is not open, so that the inverse of $c$ cannot possible be continuous? – Epsilon Away Nov 06 '21 at 12:01

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Since $(0,0)=c(0)$, you have that any open neighborhood of $c(0)$ contains $c(t)$ for $t>>1$ and $t<<-1$, $i.e.$ a small open neighborhood of $c(0)$ has 3 connected components, while a small open neighborhood of $0$ in $\mathbb R$ is connected.

ecrin
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