Making Figure-8 into a manifold So I am using the map in Lee's Smooth manifold book $\beta(t) = (\sin 2 t, \sin t)$. This is an injective immersion where $t \in (-\pi, \pi).$
Now the image is not a manifold of $\mathbb{R}^2$ because an open ball around near the origin gives a picture of "X" and that doesn't look like $\mathbb{R}$?

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*Can we make it a 1-manifold in $\mathbb{R}^2$? Or would that require the "X" to look like $\mathbb{R}$? Which is not possible.

*Is that the problem with self-intersections on $\mathbb{R}^2$? Because it would create pictures like "X" or "$\bot$" and those don't look like $\mathbb{R}$?

Next, I've been told that if we look use the induced topology by $\beta$ from $(-\pi, \pi)$, this can be made into a manifold (of $\mathbb{R}$?)
3.Even if I look near $0$ with an open interval in the image set, I am still going to get an "X" shape no? Or do they mean that if I delete the intersection point, a neighborhood near the center would "ignore" the other three pieces of the "X"?

 A: The key idea here is that switching to a different topology on the same set changes the meaning of "near" and "neighborhood".
I'll call the set of points in the image of $\beta$ by the name $\Theta$ (it's the eighth letter of the Greek alphabet and looks slightly like a numeral 8 too).
At first, we're looking at the topology induced by the standard topology of $\mathbb{R}^2$ on $\Theta$. So an open set of $\Theta$ is the intersection of  any open set of $\mathbb{R}^2$ with $\Theta$, and we can use the intersections of the open balls of $\mathbb{R}^2$ as basic neighborhoods within $\Theta$.

Now the image is not a manifold of $\mathbb{R}^2$ because an open ball around near the origin gives a picture of "X" and that doesn't look like $\mathbb{R}$?

On terminology - "A manifold of $\mathbb{R}^2$" doesn't mean much. What I'd say is if it weren't for the point $(0,0)$, $\Theta$ with this topology would be a 1-manifold embedded in $\mathbb{R}^2$. "Around near the origin" sounds like something you might not mean: Note that any point in $\Theta$ near, but not equal to, the origin is in an open set diffeomorphic to an open interval of $\mathbb{R}$. Only the origin itself is a problem. But yes, since the intersection of any open ball of $\mathbb{R}^2$ with $\Theta$ "looks like an X" and is not like a curve, there is no open set in $\Theta$ containing the origin diffeomorphic to an open interval of $\mathbb{R}$, and $\Theta$ cannot be a 1-manifold. This could be proved more formally, but that's a good intuitive idea of it.
But now when we switch to the topology induced by $\beta$, the meaning of "near" and "neighborhood" changes.

Even if I look near $0$ with an open interval in the image set, I am still going to get an "X" shape no? Or do they mean that if I delete the intersection point, a neighborhood near the center would "ignore" the other three pieces of the "X"?

There is no point deletion involved here, but we do need to pay attention to the neighborhoods used. Now an open set of $\Theta$ is the image $\beta \big((a,b)\big)$ of any open interval $(a,b) \subseteq (-\pi,\pi)$. A small open set of $\Theta$ around the origin, like $\beta\big( (-\delta,\delta) \big)$, includes only the origin and portions in the first and third quadrants, and that set is curve-like. In fact, it has an obvious chart onto $\mathbb{R}$: $\beta^{-1}$ maps this open set back to the original open interval $(-\delta,\delta) \subset \mathbb{R}$. The "tails" in the second and fourth quadrants aren't actually "near" the origin, in the sense that an open set including the origin is not forced to include them.
