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I am doing this UW Madison geometry qualifying exam, and I got stuck on this one :

Let $R=S^1\vee S^1$ as in the figure, and $X$ be the cellular 2-complex obtained by attaching two 2-cells to $R$ as indicated. Let $Y=RP^2\vee S^1,$ show that every map $f:X\rightarrow Y$ is nullhomotopic.

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It is not hard to see using algebraic means that the induced map on $\pi_1$ is the trivial map (it has been asked before but with no answer: Map to $RP^2 \vee S^1$ nullhomotopic), however because the universal cover of $RP^2\vee S^1$ is not contractible, this is not enough. Can someone please help me finish the argument please ? Thank you !

Simplyorange
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1 Answers1

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EDIT: My previous answer gave an argument which only showed that a map $X \to S^1 \vee \mathbb{R}P^3$ is nullhomotopic. Thanks to @Jason Devito for pointing it out. I planned to delete this incorrect answer, but I had some other thoughts which I offer as a mea culpa below.

One way to approach the problem is to note that the space $X$ is the cofiber (mapping cone) of a map $S^1 \vee S^1 \to S^1 \vee S^1$ given by "$(a^3, (ab)^5)$". Therefore, there is a long exact sequence of pointed sets $$\cdots \to [S^2 \vee S^2, Y] \xrightarrow{\text{(ii)}} [S^2 \vee S^2, Y] \to [X, Y] \to [S^1 \vee S^1, Y] \xrightarrow{\text{(i)}} [S^1 \vee S^1, Y].$$ In order to show that $[X,Y] = \{*\}$, we would need to show that the map labelled by (i) is injective and the map labelled by (ii) is surjective.

That (i) is injective follows from the algebraic manipulations alluded to in the OP. The map (i) takes a pair $(a, b)$ of elements in $\pi_1 Y$ to the pair $(a^3, (ab)^5)$. Since $\pi_1 Y = \mathbb{Z} * \mathbb{Z}/2$, the relation $(a^3, (ab)^5) = (1,1)$ forces $a = b = 1$, so (i) is injective.

For (ii), we have $\pi_2 Y \cong \mathbb{Z}$, but the map $\mathbb{Z}^2 \to \mathbb{Z}^2$ given by $(m,n) \mapsto (3m, 5(m+n))$ is not surjective. So this would suggest that $[X,Y] \neq \{*\}$.

JHF
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  • I don't follow one detail: a homotopy between maps from $X$ into $S^1\vee\mathbb{R}P^\infty$ can be pushed into $S^1\vee\mathbb{R}P^3$ since the domain of a homotopy is three dimensional. But why can you push the homotopy into $Y$? – Jason DeVito - on hiatus Mar 07 '21 at 23:06
  • Yes, now that I think more carefully, I have the same question. – Simplyorange Mar 07 '21 at 23:07
  • @JasonDeVito This should be via an obstruction theory argument relying on the fact $\pi_3 (\mathbb{R}P(3)) =\mathbb{Z}/2$, but this seems like a lot for a qual in geometry. – Connor Malin Mar 07 '21 at 23:23
  • @ConnerMalin: My obstruction theory is very weak, so I'm not sure why that homotopy group is relevant. (Also note that this homotopy group is $\mathbb{Z}$.) – Jason DeVito - on hiatus Mar 07 '21 at 23:28
  • @JasonDeVito Haha, got my stable and unstable confused. I was thinking this is where the fact the second cohomology has odd torsion should come in. – Connor Malin Mar 07 '21 at 23:31
  • @JasonDeVito You're right: I got my dimensions mixed up, and it's no longer clear to me if my assertion is true. I'll try to patch up the argument, and I'll delete this answer if I could not. – JHF Mar 07 '21 at 23:51