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I feel very confused about CW complex's. Mainly, taking some known space and putting a CW structure on it. For example, consider the $1$-dimensional real projective space $\mathbb{R}P^1$, lines through the origin in $\mathbb{R}^2$. The topology is given by the quotient map $q:\mathbb{R}^2\setminus\{0\}\to\mathbb{R}P^1$ sending a point to its linear span. How do I go about showing this has a cell decomposition of one $0$-cell $e^0$ and one $1$-cell $e^1$?

Based on the definition of a cell decomposition, it must be that $e^0\cup e^1=\mathbb{R}P^1$. I don't see why. But taking it as true, I then need to build a characteristic map from a closed $1$-cell $D$ to $\mathbb{R}P^1$ whose restriction to $\text{int }D$ is homeomorphic to $e^1$ and restriction to $\partial D$ is contained in $e^0$. The acclaimed characteristic map is gotten by first considering the map $F:\overline{B^1}\to\mathbb{R}^2\setminus\{0\}$ defined by $F(x)=(x,x-|1|)$, then $\Phi:\overline{B^1}\to\mathbb{R}P^1$ given by $\Phi=q\circ F$ should work. I don't see why.

I do see that $\Phi$ is a map from a closed $1$-cell to $\mathbb{R}P^1$. I see that $\Phi$ is injective, since $F$ is injective, and thus $\Phi$ restricted to $\text{int }\overline{B}^1$ is a homeomorphism onto its image, $e^1$. I see that if $x_0\in\partial\overline{B^1}$ then $F(x_0)=(x_0,0)$, so that $\Phi(x_0)=(q\circ F)(x_0)=q(x_0,0)$, but I don't see why this lands in a $0$-cell.

Rough_Manifolds
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2 Answers2

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But $\Bbb RP^1$ is just (homeomorphic to) $S^1$. This should make finding a CW structure quite a lot easier.

"Based on the definition of a cell decomposition, it must be that $e_0\cup e_1=\Bbb RP^1$. I don't see why."

I don't see why either. That's not at all "by definition". It's something that has to be shown.

I think you meant $F:x\mapsto(x,|x|-1)$. It's worth noting that $\Phi$'s restriction is a homeomorphism only follows automatically from injectivity if you have already observed that $\Bbb RP^1$ is Hausdorff. This is true, but I want to mention it anyway.

The only two $x_0\in\partial\overline{B^1}=\{-1,1\}$ are $-1$ and $1$. Because $(-1,0)$ and $(1,0)$ lie on the same line through the origin, we can notice that $\Phi(-1)=q(-1,0)=q(1,0)=\Phi(1)$, so indeed $\Phi$'s restriction to the boundary has image contained in a zero-cell - which is just a singleton - given by $e_0\to\Bbb RP^1$, $\ast\mapsto q(1,0)=q(-1,0)$.

Therefore, these maps meet the "C" - closure-finite - criterion. The target space is Hausdorff. The "W" - criterion is automatically met since you've only got finitely many cells, and the criterion that the target space is the disjoint union of the "open" cells is also met, but I recommend you take a moment to think about that.

Therefore (by the Whitehead definition / criterion) $\Phi$ together with the map $e_0\to\Bbb RP^1$ assemble to a cellulation of $\Bbb RP^1$ as $e_0\cup e_1$.

FShrike
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  • I know you are right, but I'm still not seeing how $\Phi(1)=q(1,0)$ is in the $0$-cell, and how $e^0\cup e^1=\mathbb{R}P^1$. As you say, $e^0$ is a singleton, whereas $q(1,0)$ is a line through the origin. How can these possible be the same? – Rough_Manifolds Mar 12 '23 at 15:08
  • $(1,0)$ is a point. There is an associated line through the origin, namely ${(x,0):x\in\Bbb R}$. It so happens that $(1,0)$ and $(-1,0)$ are both points on this line, and neither is zero. By very definition of $q$, we know $q(1,0)=q(-1,0)$. $q(1,0)$ isn't a line through the origin per se, it's a point in the space $\Bbb RP^1$. Let's call that point $\ell$ (it does represent a line though). However, in $\Bbb RP^1$, it is merely a point, and $\Phi(1)=\ell=\Phi(-1)$ is true as an equality of elements. The zero-cell $e_0$ is attached to $\Bbb RP^1$ by, if $e_0={\ast}$, $\ast\mapsto\ell$. – FShrike Mar 12 '23 at 15:14
  • @Rough_Manifolds Then $\Phi(\partial\overline{B^1})\subseteq e_0$ holds because both sides are equal to ${\ell}$ – FShrike Mar 12 '23 at 15:15
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    Ah, I see this all stems from my misunderstanding of what $\mathbb{R}P^n$ really is. Thanks! – Rough_Manifolds Mar 12 '23 at 15:17
  • @Rough_Manifolds You're welcome. $\Bbb RP^n$ is "just" a quotient space. I used to have a hard time managing quotients in topology, they are somehow more confusing than quotients in algebra – FShrike Mar 12 '23 at 15:18
  • I agree! I still can't get why $e^0\cup e^1=\mathbb{R}P^1$, however. Why should this be true? – Rough_Manifolds Mar 12 '23 at 15:21
  • @Rough_Manifolds Do you mean, why is $e_0\cup e_1=\Bbb RP^1$ true in the sense that, "why should the image $\Phi([-1,1])$ cover all of $\Bbb RP^1$?" Or in the sense that, "why is this a valid cellulation?" – FShrike Mar 12 '23 at 15:23
  • Where I still feel confused is why can't we say there are more cells, such as another $0$-cell, or $1$-cell, etc. What is stopping us from saying this? – Rough_Manifolds Mar 12 '23 at 15:32
  • @Rough_Manifolds Absolutely nothing is stopping us from saying this. There is no one, unique way to cellulate a space. We could cellulate $\Bbb RP^1$ with many many many cells, but the most efficient way is to do it as $e_0\cup e_1$ – FShrike Mar 12 '23 at 15:33
  • That makes sense. So, how do we know when we have cellulated a space? For example, why does just $e_0$ not cellulate $\mathbb{R}P^1$ but $e_0$ together with $e_1$ does? – Rough_Manifolds Mar 12 '23 at 15:37
  • @Rough_Manifolds $e_0$ doesn't cellulate $\Bbb RP^1$ because it fails the most basic criterion: it doesn't cover the target space by disjoint cells. $\Bbb RP^1$ is not a one-point space! A more subtle case where something covers the target space but isn't a valid cell complex is the case where we try and fail to cellulate the closed unit disk $D^2\subset\Bbb R^2$ by $S^1$-many zero-cells $e_0$, which for every $\theta\in S^1$ are simply attached to $\theta\in D^2$, and one two-cell $e_2$ which is attached by the identity map $D^2\to D^2$ – FShrike Mar 12 '23 at 17:26
  • This satisfies the disjoint covering criterion, the "W" axiom, and indeed $\partial e_2$ is covered by cells of lower dimension... but not by finitely many, so this violates the "C" criterion in "CW" – FShrike Mar 12 '23 at 17:27
  • Thank you! It took some time for me to digests the concept, but I get it now. – Rough_Manifolds Mar 13 '23 at 01:38
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How do I go about showing this has a cell decomposition of one $0$-cell $e^0$ and one $1$-cell $e^1$?

I guess you somewhere got the information that this is true. If you assume it as true, then in fact $e^0 \cup e^1$ must be a copy of $S^1$ because there is only one map $\partial D^1 = S^0 \to e^0$ so that $e^0 \cup e^1 = D^1/S^0 \approx S^1$.

Thus a strategy could be to prove that $\mathbb RP^1 \approx S^1$. This is easy; see for example Help understand the construction of the real projective line $\mathbb{R}P^1$.

A more systematic approach is this:

It is well-known that $\mathbb RP^1$ can alternatively be defined as the quotient of $S^1$ by identifying all pairs $z, -z$ of antipodal points. For a formal proof see When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map?

Now obviously $S^1 \subset \mathbb R^2$ has a CW-structure given by

  • Two $0$-cells $d^0_\pm = \{ (\pm 1,0) \}$.

  • Two open $1$-cells $d^1_\pm = \{ (x,y) \mid x^2 + y^2 = 1, (-1)^{\pm 1} y > 0 \}$.

Characterstic maps for $d^1_\pm$ are $\phi^1_\pm : D^1 \to S^1, \phi^1_\pm(x) = (x, \pm \sqrt{1 - x^2})$.

For $i = 0,1$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_\pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1$ form an open cell decomposition of $\mathbb RP^1$. The characterstic map for $e^1$ is induced by those for the $1$-cells of $S^1$, i.e. we have $\psi^1 = p\phi^1_\pm$.

Paul Frost
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