Constructing a homeomorphism from 2 spaces commonly used to define $\mathbb{RP}^2$ I’m playing around with a book called “Topology, a categorical approach” (by Tai-Danae Bradley, Tyler Bryson, John Terilla) and early on a small exercise shows up which I can sort of understand intuitively but it’s quite unclear how to go about proving what’s asked.
The books gives two different definitions of $\mathbb{RP}^n$. The first one is the quotient of $\mathbb{R}^{n+1}\setminus\{0\}$ by the equivalence relation $x\sim \lambda x$, $\lambda\in\mathbb{R}$.
The second definition is given just for $\mathbb{RP}^2$, by identifying sides of $I^2$ (where $I=[0,1]$). That is, $\mathbb{RP}^2$ is defined as $I^2/\sim$ where $(x,0)\sim (1-x,1)$ and $(0,y)\sim(1,1-y)$.
The question is then to show that the spaces obtained by the two constructions are indeed homeomorphic. Note that at this point the books has only introduced the very basic definitions of point-set topology, but with three perspective (standard textbook definition, somewhat “better” definition, categorical phrasing using some universal property) and also some categorical notions because that’s kind of the point of the book.
So I’m trying to find the “simplest” (ie most low tech/ elementary) way to show the spaces are indeed homeomorphic but I’m kind of stuck. I don’t really know where to go if that makes sense. Any help is appreciated. I think I’m not really sure how to think about $\mathbb{RP}^2$ as a topological space whose points are lines through the origin in $\mathbb{R}^3$, not sure what to make of it. This feels like it should be “an easy verification” but I don’t see it right off the bat.
 A: The simplest approach is to show that there are a set of equivalence class representatives in the "lines through the origin" version that are homeomorphic to the square and have the same identifications called out for the square.
Let's first think about the unit sphere, $S^2$ in $\Bbb{R}^3$.  Every line through the origin intersects this sphere twice in its antipodes.  So already, we have picked two points on each line (two representatives from each equivalence class of points on a line through the origin).  It can still be mentally taxing to make the final identification of every pair of antipodes on the sphere.  So instead, think of the closed upper half-sphere: $\{(x,y,z) \mid x^2 + y^2 + z^2 = 1, z \geq 0\}$.  Its interior is single representatives from the equivalence classes that don't lie in the $xy$-plane.  This is a disk (Project the $z$ coordinate to $0$ to see this immediately.) with identifications on its sides.  In particular, any point $(x,y,0)$ is identified with $(-x,-y,0)$ since those two points lie in the same equivalence class.
Break the boundary up by the usual four quadrants of the $xy$-plane.  In each quadrant, notice the identifications are to the opposite quadrant.  Moving in the positive (increasing angle) direction through the first quadrant, the identified point in the third quadrant is also moving in the positive direction -- exactly as if there were a half-twist between the points of these two quarter-circles.  The same happens in quadrants two and four.
Now switch to the square.  It's easy enough to see that it is homeomorphic to a disk with identifications along its boundary.  We can arrange for that disk to be the closed upper unit hemisphere with the four original sides mapped one to each quadrant of the $xy$-plane.  The edge that has been mapped to each quadrant is identified to the opposite quadrant, with a half-twist.
Now verify that the identifications carried along on the boundary of the square are actually the same as for the hemispherical equivalence classes.  (If not, it means we mapped the square "upside-down".  Flip it over ($x \mapsto -x, y \mapsto y$) so the identifications run around the circumference in the other direction.)
Summary: both descriptions can be presented as a disk with boundary with the same quotient along its boundary.
