I was trying to figure out how to construct homeomorphisms using the universal quotient property, which I didn't know about until now. And this nicely gives the criteria for doing this. However is there any way I can substitute numbers $(3)$ and $(4)$ given below by either showing the function $f$ is surjective or $g$ is injective and surjective, instead of finding an inverse function which seems difficult? Also would $(2)$ be equivalent to showing that $\{f^{-1}(y)\}$ gives the partition of $X / \sim$?
For example I was trying to use this to show the real projective plane $\mathbb{R}P^2$ is homeomorphic to $S^2/(x \sim -x)$. I already did this problem. Could someone explain if this is the right approach.
Define $f:S^2 \rightarrow \mathbb{R}P^2$ by $f(x)=tx$ where $t \in \mathbb{R}$ is a parameter.
Then $x \sim y \implies \text{span}\{x\}=\text{span}\{y\} \implies tx=ty$. So by the Universal property of the quotient topology, f is continuous and there exists a unique function $g:S^2 /(x \sim -x) \rightarrow \mathbb{R}P^2$ That is well defined and continuous. $g$ is bijective since $g([x])=g([y]) \implies tx=ty \implies [x]=[y]$. $g$ is surjective since for $tx \in \mathbb{R}P^2$, set $\tilde{x}=\frac{x}{||x||}$ then $g([\tilde{x}])=tx$. So $f$ induces a homoemorphism $g$.
Also how would this work for proving a homeomorphism between spaces where an equivalence relation is defined by equivalence classes as sets?
For example how could I generalize this approach for this problem?
Best Answer
Yes 2 is equivalent to $f^{-1}[\{y\}], y \in Y$ being equal to the partition of $X$ into classes. This both shows 1-1-ness of $\tilde{f}$ (if two classes map to the same $y$ under $\tilde{f}$, they were already in the same class) as well as well-definedness (if two points are from the same class, they have the same $f$ image).
In your $\Bbb S^2, \mathbb{ RP}^2$ example the universal property of the topology implies that if $f$ is continuous, so is $g$ (as $g = f \circ q$), so the work is in showing $f$ continuous. $2$ is easy as two distinct points on the sphere define the same line iff they are antipodal. That fact, plus that every class is obtained, makes $g$ a bijection. You still have the show continuity of $f$ though.
Note that in this case the last two steps are unnecessary as a continuous bijection from a compact space to a Hausdorff space is already a homeomorphism.