2

enter image description here

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?

Show torus homeomorphic to $S^1 \times S^1$

ernesto
  • 549

1 Answers1

1

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.

Henno Brandsma
  • 242,131
  • Does $2$ imply the continuity of $f$? To where I do not have to show it directly? In general can I show bijectiveness by showing $g({x,-x})=g({y,-y}) \implies {x,-x}={y,-y}$? And show it is surjective by letting $tx \in \mathbb{R}P^2$ and $g{\frac{x}{||x||},-\frac{x}{||x||}}=tx$? – ernesto May 13 '21 at 06:13
  • @ernesto $2$ implies we can apply the universal property to show continuity of $\tilde{f}$ once we’ve proved the continuity of $f$ first. That map should be continuous from the beginning. – Henno Brandsma May 13 '21 at 06:15
  • Is there any reason I shouldn't define homeomorphisms on the equivalence classes explicitly like in this problem I tried? https://math.stackexchange.com/questions/4125054/show-torus-homeomorphic-to-s1-times-s1 – ernesto May 13 '21 at 06:22
  • Also I defined the homeomorphisms on the equivalence classes of the identification space explicitly in this problem https://math.stackexchange.com/questions/4126329/real-projective-plane-isomorphic-to-quotient-space-of-sphere Is there a reason I shouldn't do this? – ernesto May 13 '21 at 06:29
  • @ernesto I wrote an answer to the torus question, where the continuity of the starting function is more clear. Here the imega of a point on $\Bbb S^2$ is itself a class, so the continuity must be checked in $\Bbb R^2$, depending on your definition of $\mathbb{RP}^2$, what is that definition? – Henno Brandsma May 13 '21 at 07:34