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We define a space $X$ by $$X=(S^2\times S^2)\cup_{\Delta} D^3$$ where $S^2$ is the $2$-sphere, $D^3$ is the $3$-disk, and $\Delta\colon S^2\to S^2\times S^2$ is the diagonal map, so we attach a $3$-cell to $S^2\times S^2$ using the diagonal.

I want to compute the (co)homology groups of this space, but I don't know how. I would say that this space is not a manifold since dimensions don't match, and it doesn't carry a CW structure, so I'm a bit lost. What do you think it would be the best strategy to solve this problem?

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    $\begingroup$ So the diagonal element is getting killed... What element the diagonal represents in cohomology of $S^2\times S^2$? $\endgroup$ – Anubhav Mukherjee May 22 '18 at 14:40
  • $\begingroup$ You can make this a CW complex. $\endgroup$ – Cheerful Parsnip May 22 '18 at 17:23
  • $\begingroup$ You might try visualizing what happens for $S^1\times S^1\cup_\Delta D^2$ as a warm-up. $\endgroup$ – Cheerful Parsnip May 22 '18 at 17:25
  • $\begingroup$ How can I make this space a CW complex? $\endgroup$ – Laszlo May 23 '18 at 0:46
  • $\begingroup$ @Laszlo: you have to subdivide the existing cells. That's why I suggested looking at one dimension down as a warm-up to see what needs to be done. $\endgroup$ – Cheerful Parsnip May 23 '18 at 19:28
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I'll sketch an argument using the Mayer-Vietoris sequence, which is a little more low-tech than Tyrone's answer, and fleshes out a comment by Anubhav Mukherjee. Let $U$ be a small open neighborhood of $S^2\times S^2$ in your adjunction space $X$ and let $V$ be the interior of the attached $D^3$. Then we have that $H_i(D^3)$ is trivial (let's take reduced homology). Also $U\cap V$ deformation retracts onto the diagonal $S^2$ inside $S^2\times S^2$. Our sequence becomes $$\to H_i(U\cap V)\to H_i(S^2\times S^2)\to H_i(X)\to H_{i-1}(U\cap V)\to$$ Since $H_i(U\cap V)=0$ unless $i=2$, we get that $H_k(S^2\times S^2)\to H_k(X)$ is an isomorphism for $k=1,3,4$. In the middle of the sequence we have $$0\to H_2(U\cap V)\to H_2(S^2\times S^2)\to H_2(X)\to 0.$$ Moreover, we explicitly know the map is induced by the diagonal so we can set things up so that $\mathbb Z\to\mathbb Z\oplus \mathbb Z$ by the diagonal $n\mapsto (n,n)$. So $H_2(X)\cong(\mathbb Z\oplus \mathbb Z)/\langle (n,n)\rangle\cong \mathbb Z$.

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We have homotopy cofibration sequences

$S^2\xrightarrow{c}S^2\vee S^2\xrightarrow{(1,-1)}S^2$

$S^3\xrightarrow{w}S^2\vee S^2\xrightarrow{j}S^2\times S^2$

where $c$ is the suspension comultiplication, $w$ is the attaching map of the top cell of the product, and $j$ is the inclusion.

Now $\Delta\simeq j\circ c$, so if we intertwine the previous homotopy cofibration sequences then we find that $X$ has the homotopy type

$X\simeq (S^2\times S^1)\cup_\Delta e^e\simeq S^2\cup_{[1,-1]} e^4$

where $[1,-1]=(1,1)\circ w$ is the Whitehead product. Now the Whitehead product is bilinear, so $[1,-1]=-[1,1]$. Moreover it is well known that $[1,1]=-2\eta$, where $\eta:S^3\rightarrow S^2$ is the Hopf map that generates $\pi_3S^2$. Hence $[1,-1]=2\eta$ and

$X\simeq S^2\cup_{2\eta}e^4.$

We consider cohomology. Thus

$H^nX=\begin{cases}\mathbb{Z}&n=0\\\mathbb{Z}\{x\}&n=2\\\mathbb{Z}\{y\}&n=4\\ 0&\text{otherwise}\end{cases}$

where $x$, $y$ denote generators. From the fact that $\eta$ is the map of Hopf invariant one we conclude that

$x^2=4y$

(you may also want to recall that $\eta$ is the attaching map of the top cell of $\mathbb{C}P^2$).

This calculates the ring structure of $H^*X$. Since the groups are free you can dualise with the universal coefficient theorem to obtain the homology, which I will leave up to you.

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  • $\begingroup$ This answer seems very complete but it is too advanced for me. I haven't seen the Whitehead product, nor homotopy cofibrations. $\endgroup$ – Laszlo May 22 '18 at 17:20
  • $\begingroup$ I was worried that might be the case, but wanted to show that the general problem of the space's homotopy type isn't really much more difficult that its homology. $\endgroup$ – Tyrone May 22 '18 at 17:28
  • $\begingroup$ I appreciate your effort -- I will try to understand it if I get a bit of time. $\endgroup$ – Laszlo May 22 '18 at 17:32
  • $\begingroup$ You can essentially take the above as the definition of the Whitehead product. If $f:S^n\rightarrow X$, $g:S^m\rightarrow X$ are maps, then let $w:S^{n+m-1}\rightarrow S^n\vee S^m$ be the attaching map for the top cell of $S^n\times S^m$. Then the composition $[f,g]=(f,g)\circ w:S^{n+m-1}\rightarrow X$ is the Whitehead product of $f$ and $g$. $\endgroup$ – Tyrone May 22 '18 at 17:33
  • $\begingroup$ As for "homotopy cofibration" it essentially means here little more than a cofibration sequence. Each map is not necessarily a cofibration, but if you turn it into one then all the spaces have the right homotopy type. $\endgroup$ – Tyrone May 22 '18 at 17:36

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