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Define a quotient topology on $\mathbb{R}^2$ such that the result is homeomorphic to a sphere and second to a closed rectangle with all the interior points included.

For the first one: I know that $\mathbb{R}^2$ is homeomorphic to a sphere (without the nort pole). But I don't know how to get 'some extra point' with defining a quotient topology. I would rather say you get 'less points' by identifying some points.

For the second: You have to identify all points outside some rectangle? Then you get a rectangle and one point that identifies all points outside the rectangle? Is that the result?

Thanks in advance

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    Your second solution does not lead to a rectangle. If you identify all points outside the rectangle as one point, then this point is in any open neighborhood of the points on the boundary of the rectangle. I would find a quotient of $\mathbb{R}^2$ that is isomorphic to a closed disc, then because a closed disc is isomorphic to a closed rectangle, we are done. – 2017-01-28

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Your second solution is actually very close to the answer for the first problem: identify all points outside some open disc to each other: the open disc is homeomorphic to a sphere minus the north pole, and the identification makes all the rest of the plane into said north pole.

As for the rectangle, draw the rectangle in the plane, and identify each point outside the rectangle with the point on the boundary of the rectangle that it is closest to.

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    Ok, Thanks a lot!! In the case of a quotient topology on $\mathbb{R}^2$, s.t.the result is homeomorphic to say a line through the origin. It suffices to consider vertical lines and then identify all points on that line, with the one of the intersection? – 2017-01-28
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    @bob Exactly. Although there are other ways of doing it, that is probably the simplest. – 2017-01-28
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    Ok, Thanks for the answer! – 2017-01-28
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To get a sphere, you can take an open ball $B(0,r) \subseteq \mathbb{R}^2$ and collapse all the points in the complment $$ \mathbb{R}^2 \setminus B(0,r) = \{ x \in \mathbb{R}^2 \, | \, \| x \| \geq r \}$$ to a single point. This has the same effect as adding on extra point to $B(0,r)$ and since $B(0,r)$ is homeomorphic to $\mathbb{R}^2$, you'll get the sphere.

Instead of a closed rectangle, it might be more intuitive to think how you can get a closed ball. To do that, start with an open ball $B(0,1)$ and identify each ray $t (x,y)$ (where $x^2 + y^2 = 1$ and $t \geq 1$) with the point $(x,y)$ on the boundary. This way, you collapse each ray to a single point on the boundary and get a closed ball. Since a closed ball is homeomorhic to a closed rectangle, you also get a solution to the original question. Alternatively, you can try and mimic this idea for the closed rectangle (only in this case it is less obvious what are the "rays" you want to identify with the boundary points of the rectangle).

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    Ok, Thanks a lot! – 2017-01-28
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    Just a question: The quotient topology on $\mathbb{R}^2$, then corresponds with the subspace topology on the sphere, right? Opens around the northpole, correspond than to $\mathbb{R}^2\backslash B(0,r)$ union an open in the disc around the boundary. Is that correct or not? – 2017-01-28
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    Yep, so for example $\mathbb{R}^2 \setminus \overline{B(0,\frac{r}{2})}$ would correspond to an open neighborhood around the north pole. – 2017-01-28
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    Yes indeed, thanks again! – 2017-01-28