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$\mathbf {The \ Problem \ is}:$ Let $w: S^{3} \rightarrow S^{2} \vee S^{2}$ be the attaching map of the 4-cell of $S^{2} \times S^{2}$ (using the usual CW complex structiure on $S^{2} \times S^{2}$ ). Let $\nu: S^{2} \vee S^{2} \rightarrow S^{2}$ be the fold map (the map which is identity on each factor). Let $J$ be the mapping cone of $\nu \circ w$. Compute $H^{*}(J)$ as a ring.

$\mathbf {My \ approach}:$ I have attached an image what I have tried . Now can we complete the commutative diagram in some way which will imply naturality of long exact sequence in cohomology and then we can approach(as upper and lower rows are co-fibration) ?

And how to find effect of $\nu$ in cohomology ? Is it $a \mapsto (a,a)$ where $a$ generates $H^2(S^2) ?$

Can anyone explicitly describe the cup product? Thanks in advance .

enter image description here

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  • $\begingroup$ Ignoring $\mathbb{C}P^2$ for a moment, have you tried comparing the cofibration sequence defining $J$ with that defining $S^2\times S^2$? You need to know the action of the fold map in cohomology, and your guess here is correct (reason: the isomorphism $H^2(S^2\vee S^2)\xrightarrow\cong H^2S^2\oplus H^2S^2$ is effected by summing together the maps induced by the two inclusions of $S^2$ as the summands of $S^2\vee S^2$). $\endgroup$
    – Tyrone
    Apr 20 at 17:06
  • $\begingroup$ By defining $S^2 × S^2$, do you mean $S^3 \to S^2 \vee S^2 \hookrightarrow S^2 × S^2$ where the first map is $w ?$ $\endgroup$ Apr 20 at 18:10
  • $\begingroup$ Yes. The folding map $S^2\vee S^2\rightarrow S^2$ now induces $S^2\times S^2\rightarrow J$. $\endgroup$
    – Tyrone
    Apr 20 at 18:16
  • $\begingroup$ Can you please post an answer ? It will be very helpful for me . $\endgroup$ Apr 20 at 18:16

1 Answer 1

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There are cofibration sequences $$S^3\xrightarrow\omega S^2\vee S^2\xrightarrow{i} S^2\times S^2$$ $$S^3\xrightarrow\theta S^2\xrightarrow{j} J$$ where $\omega$ attaches the top cell of $S^2\times S^2$, the map $\theta$ is the composition $\theta=\nu\circ \omega$, with $\nu:S^2\vee S^2\rightarrow S^2$ being the fold map, $J=S^2\cup_\theta e^4$, and $i,j$ are the inclusions.

Since $\nu(\omega(x))=\theta(x)$, there is an induced map $\varphi:S^2\times S^2\rightarrow J$ which is $\nu$ on $S^2\vee S^2$ and is the identity on the points of the open 4-cell. In particular, if $p:S^2\times S^2\rightarrow S^4$ is the map collapsing the 2-skeleton to a point, then $p$ factors as the composition $$p:S^2\times S^2\xrightarrow{\varphi}J\xrightarrow{q}S^4$$ where $q:J\rightarrow S^4$ is the corresponding collapse map on $J$.

Now turn to cohomology. I will use $H^*$ to denote reduced singular homology with integral coefficients.

The maps $\omega,\theta$ induce trivial maps in cohomology for degree reasons, and this leaves short exact sequences $$0\leftarrow H^*(S^2\vee S^2)\xleftarrow{i^*} H^*(S^2\times S^2)\xleftarrow{p^*} H^*S^4\leftarrow 0$$ $$0\leftarrow H^*S^2\xleftarrow{j^*} H^*J\xleftarrow{q^*} H^*S^4\leftarrow 0.$$

Here we have used the fact that $H^*(J,S^2)\cong H^*(J/S^2)=H^*S^4$ naturally, and similarly $H^*(S^2\times S^2,S^2\vee S^2)\cong H^*S^4$ naturally. There is a map between these sequences (going upwards), which is $\nu^*$ on the left-hand group, $\varphi^*$ on the middle group, and the identity on $H^*S^4$.

Let $s\in H^2S^2\cong\mathbb{Z}$ and $t\in H^4S^4\cong\mathbb{Z}$ be generators. By exactness we have $H^2J\cong\mathbb{Z}$ generated by a class $x$ satisfying $j^*x=s$, and $H^4J\cong\mathbb{Z}$ generated by $y=q^*t$. Of course $H^kJ=0$ for $k\not\in\{2,4\}$. Thus the problem facing us is reduced to the computation of the integer $n\in\mathbb{Z}$ satisfying $$x^2=n\cdot y.$$

Now, the maps $$in_1:S^2\hookrightarrow S^2\vee S^2\hookleftarrow S^2:in_2$$ which include the two wedge summands add together to induce an isomorphism $$H^*(S^2\vee S^2)\xrightarrow{\cong} H^*S^2\oplus H^*S^2$$

Since $\nu\circ in_1=id_{S^2}=\nu\circ in_2$, the composition of this isomorphism with $\nu^*:H^*S^2\rightarrow H^*(S^2\vee S^2)$ is exactly the map $$a\mapsto (a,a).$$

This map is injective, so use of the Five Lemma shows that $\varphi^*:H^*J\rightarrow H^*(S^2\times S^2)$ is injective.

Next use the Kunneth formula to identify generators of $H^2(S^2\times S^2)\cong\mathbb{Z}\oplus\mathbb{Z}$ as the classes $s_1=s\otimes 1$ and $s_2=1\otimes s$. Observe that $$i^*\varphi^*(x)=\nu^*j^*(x)=\nu^*s=(s,s).$$ Since $i^*:H^2(S^2\times S^2)\rightarrow H^2(S^2\vee S^2)\cong H^2S^2\oplus H^2S^2$ is an isomorphism we have $$\varphi^*x=s_1+s_2.$$ From this we obtain $$\varphi^*(x^2)=(\varphi^*x)^2=(s_1+s_2)^2=s_1^2+s_1s_2+s_2s_1+s_2^2=2s_1s_2,$$ since $s_1^2=s_2^2=0$ and the even degree classes satisfy $s_1s_2=s_2s_1$.

Now the class $s_1s_2=s\otimes s$ generates $H^4(S^2\times S^2)$, and we can assume that the generator $t\in H^4S^4$ satisfies $p^*t=s_1s_2$. Thus $$\varphi^*y=\varphi^*q^*(t)=p^*(t)=s_1s_2.$$ Comparing this with the equation above we have $$\varphi^*(x^2)=2s_1s_2=2\cdot \varphi^*(y)=\varphi^*(2\cdot y).$$

But $\varphi^*$ is injective, so the only way this equation can hold is if $$x^2=2\cdot y,$$ which is exactly what was to be shown.

Incidentally, $\theta=\nu\circ\omega$ must be homotopic to a multiple of the Hopf map $\eta$ which generates $\pi_3S^2\cong\mathbb{Z}$. Our computation above verifies that $\theta\simeq\pm 2\cdot\eta$.

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