# Exercises:Saul - Algebraic Topology - 9/Exercise 9.7

## Exercises

### Exercise 9.7

Show that the cone on the real projective plane, i.e. [ilmath]C(\mathbb{RP}^2)[/ilmath], is not a topological 3-manifold with boundary.

I might have two solutions for this, I accidentally answered the wrong question initially. [ilmath]\newcommand{\crp}{C(\mathbb{RP}^2)} [/ilmath]

#### Solution 1

I struggled to focus writing this up, however the idea should be clear, if the presentation isn't up to my usual standard please forgive me. Aditionally:
• [ilmath]\langle G\rangle[/ilmath] means the "normal subgroup generated by [ilmath]G[/ilmath]", fortunately where it is used here [ilmath]G[/ilmath] is trivial, thus is a normal subgroup itself so [ilmath]\langle 0\rangle\eq 0[/ilmath], thus the question of "normal subgroup closure" never pops up in this solution.

Before we start, I want to note that we make heavy use of the fundamental group, and I may write this without a base-point, this is because for a path-connected space the fundamental group based anywhere is isomorphic to any other fundamental group based elsewhere of the space. Please know that I know this.

First we consider [ilmath]\crp[/ilmath] as a topological space and work out some fundamental group isomorphisms.

• Immediately, as any topological cone is contractible space we see:
• [ilmath]\pi_1(\crp)\cong 0[/ilmath] (where [ilmath]0[/ilmath] denotes the trivial group, we do not use [ilmath]1[/ilmath] even though fundamental groups are usually not Abelian)

Let [ilmath]p\in\crp[/ilmath] be the apex of the cone, i.e. the image of [ilmath]\mathbb{RP}^2\times\{1\} [/ilmath] under the the cone's quotient map, some people may identify [ilmath]X\times\{0\} [/ilmath] as the apex, we do not.

Consider [ilmath]\crp-\{p\} [/ilmath], this is easily seen to be homeomorphic to [ilmath]\mathbb{RP}^2\times[0,1)[/ilmath]

• Note that [ilmath]\mathbb{RP}^2\times[0,1)[/ilmath] has [ilmath]\mathbb{RP}^2[/ilmath] as a strong deformation retraction (by slowly "projecting down" along [ilmath][0,1)[/ilmath]), and thus:
• [ilmath]\pi_1(\crp-\{p\})\cong\pi_1(\mathbb{RP}^2)\cong\frac{\mathbb{Z} }{2\mathbb{Z} } [/ilmath]

We summarise:

• [ilmath]\pi_1(\crp)\cong 0[/ilmath] and
• [ilmath]\pi_1(\crp-\{p\})\cong\pi_1(\mathbb{RP}^2)\cong\frac{\mathbb{Z} }{2\mathbb{Z} } [/ilmath]

We now use the "fact" it is a manifold to tackle the claim: [ilmath]\pi_1(\crp)\cong\pi_1(\crp-\{p\})[/ilmath] - this will make use of the Seifert-Van Kampen theorem.

• Let [ilmath]x\in\crp[/ilmath] be an arbitrary point that is not the implicit basepoint for [ilmath]\pi_1[/ilmath]
1. suppose that [ilmath]x[/ilmath] is in the interior of the manifold[Note 1], then:
2. now suppose [ilmath]x\in\partial(\crp)[/ilmath] - that is [ilmath]x[/ilmath] is a boundary point of our 3-manifold with boundary[Note 3]
• We see that there exists a homeomorphism: [ilmath]\varphi:U\rightarrow\mathbb{H}^3[/ilmath] where [ilmath]U[/ilmath] is an open neighbourhood of [ilmath]x[/ilmath] and [ilmath]\mathbb{H}^3[/ilmath] denotes the 3-dimensional half space:
• [ilmath]\mathbb{H}^3:\eq\{(x_1,x_2,x_3)\in\mathbb{R}^3\ \big\vert\ \Vert (x_1,x_2,x_3)\Vert < 1 \wedge x_3\ge 0\} \eq \mathbb{B}^3\cap\{(x_1,x_2,x_3)\in\mathbb{R}^3\ \big\vert\ x_3\ge 0\} [/ilmath]
• Such that [ilmath]\big(\varphi(x)\big)_3\eq 0[/ilmath] (i.e. on the [ilmath]x_3\eq 0[/ilmath] plane in [ilmath]\mathbb{R}^3[/ilmath])
• Again WLOG we can suppose [ilmath]\varphi(x)\eq 0[/ilmath] - for the same reasoning as the previous case[Note 4]
• As before we note that [ilmath](U-\{x\})\cong(\mathbb{H}^3-\{0\})[/ilmath] by the restriction of [ilmath]\varphi[/ilmath] to the left hand side of the [ilmath]\cong[/ilmath]
• We use the Seifert-Van Kapmen theorem again:
• $\pi_1(\crp)\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(U)}{\langle \pi_1\big((\crp-\{x\})\cap U\big)\rangle}$[Note 5] which by homeomorphic spaces have isomorphic fundamental groups:
$\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(\mathbb{H}^3)}{\langle\pi_1(U-\{x\})\rangle}$ (by replacing spaces with homeomorphic ones)
$\cong\frac{\pi_1(\crp-\{x\})\ast\pi_1(\mathbb{H}^3)}{\langle\pi_1(\mathbb{H}^3-\{0\})\rangle}$ (by replacing spaces with homeomorphic ones)
$\cong\frac{\pi_1(\crp-\{x\})\ast 0}{0}$ as [ilmath]\mathbb{H}^3[/ilmath] and [ilmath]\mathbb{H}^3-\{0\} [/ilmath] are both clearly contractible. (To convince yourself of this take the origin for [ilmath]\mathbb{H}^3[/ilmath], you can just "straight-line" pull everything onto it. For [ilmath]\mathbb{H}^3-\{0\} [/ilmath] pick the point [ilmath](0,0,1)[/ilmath] to pull back on, again straight line onto this point)
[ilmath]\cong\pi_1(\crp-\{x\})[/ilmath]
• We conclude:
• Thus in either case:
• [ilmath]\pi_1(\crp)\cong\pi_1(\crp-\{x\})[/ilmath]
• Since [ilmath]x\in\crp[/ilmath] was arbitrary (and not the implicit base-point) we see [ilmath]\pi_1(\crp)\cong\pi_1(\crp-\{x\})[/ilmath] holds for all [ilmath]x[/ilmath]
• Next we combine everything to see:
• [ilmath]0\cong\pi_1(\crp)\cong \pi_1(\crp-\{p\})\cong\pi_1(\mathbb{RP}^2)\cong\frac{\mathbb{Z} }{2\mathbb{Z} } [/ilmath]
• Thus: [ilmath]0\cong\frac{\mathbb{Z} }{2\mathbb{Z} } [/ilmath] - obviously a contradiction!
• Thus [ilmath]\crp[/ilmath] cannot be a [ilmath]3[/ilmath]-manifold with boundary
• In fact it cannot be a [ilmath]3[/ilmath]-manifold without boundary either, as per the first half where we assumed [ilmath]x[/ilmath] was a point interior to the manifold.

This completes the proof.