Projective plane homeomorphic to a cone

I'm studyig algebraic topology and I am having several issues when comes to work out the exercises, in this case, the problem is from Bredon's Topology and Geometry. The statement of the exercise is: Show that the projective plane $\mathbb R\mathbb P^2$ is homeomorphic to the mapping cone of the map $z \rightarrow z^2$ of the unit circle in the complex number to itself.

So if I define $f:S^1\rightarrow S^1$ with $f(z)=z^2$, I know that the cone is $M_f/(S^1\times\{1\})$, where $M_f=S^1\times [0,1]\sqcup_{f_0}S^1$, $M_f$ is the cilinder, and $\sqcup_{f_0}$ means the equivalence relation $f(z)\sim (z,0)$ in the disjoint union.

So in this particular example I see that the map $f$ twist the upper circle when making the cilender, and complete two turns over the circle (for example I saw what happens to the points $(1,0)\sim 1$, $(-1,0)\sim 1$, $(i,0)\sim -1$, and $(-i,0)\sim -1$).

After this deformation (the twist part), to get the cone I just glue the top part (where I kept the points as they were) all together into a single point. So what I get is this weird space, and I can't even imagine how to find an homeomorphism from this to $\mathbb R\mathbb P^2$.

Any tips or how to look for this homeomorphism?

• What's your definition of $\mathbb{R}P^2$? (It's perfectly fine to define it as this mapping cone.) – JHF Aug 30 '17 at 19:23
• Ok, I didnt know this definition, but I use the most usual, as $\mathbb R\setminus \{0\}/\sim$ where $x\sim y$ if there is a $\lambda \in \mathbb R\setminus \{0\}$ such that $x=\lambda y$. I also have proved the same relation o the sphere $S^2$, and the relation of antipodals of the boundary of $D^2$. – e.turatti Aug 31 '17 at 0:35

OK, here's how you can construct the map. Since you've already shown that $\mathbb{R}P^2 \cong D^2/{\sim}$ where the antipodal points on the boundary are identified, I will produce a map $$C_f = (S^1 \times [0,1] \sqcup_{f_0} S^1)/(S^1 \times \{1\}) \to D^2/{\sim}.$$

First, define a map $S^1 \times [0,1] \to D^2$ by $$(\theta, r) \mapsto (1-r)e^{i\theta}.$$ Composition with the quotient map $D^2 \to D^2/{\sim}$ gives a map $S^1 \times [0,1] \to D^2/{\sim}$.

Second, define a map $S^1 \to D^2/{\sim}$ by $\theta \mapsto e^{i\theta/2}$. Ordinarily, this would not be a well-defined function from the circle to the disk $D^2$. Why is it OK in this case?

Now I can glue these two functions together to get a function $$M_f = S^1 \times [0,1] \sqcup_{f_0} S^1 \to D^2/{\sim}.$$ You should check that the points identified by $f_0$ are sent to the same point in the target, so that this map is indeed well-defined.

Next, observe that the points in $S^1 \times \{1\}$ are collapsed to a point. So the map above factors through $$M_f/(S^1 \times \{1\}) = C_f \to D^2/{\sim}.$$ This is your desired map.

A quick way to show that this is a homeomorphism is to recall that a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism. So you just need to just the map just constructed is a bijection. It's clear that this map is surjective, and checking injectivity requires a bit more work which you should really work out yourself, because it explains some of the possibly opaque choices I've made along the way.

So there are two ways you should do after reading this answer. First, do verify that the map is well-defined (I've pointed out the places that need checking) and show the last bit about the injectivity. Second, try to visualize what is happening geometrically -- this is valuable intuition that may help you later on -- and see whether you could have come up with this answer yourself now that you know what is happening geometrically. Write back if you want more clarification!

The projective plane can be thought of as the northern hemisphere with opposite points of the equator identified. The northern hemisphere is a disk and the degree 2 map on its boundary precisely achieves the identification of "opposite" points, so you get your mapping cone almost for free.