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In section $5.1$ of Hatcher's note about spectral sequences, he starts to compute stable homotopy group $\pi_{n+k}(S^{n}),k \leq 3$. Particularly for $\pi_{n+1}(S^{n})$, by Freudenthal suspension theorem it's enough to compute $\pi_{4}(S^{3})$. A part of Postnikov tower for $S^{3}$ is: $$K(\pi_{4}S^{3},4) \to X_{4} \to K(\mathbb{Z},3)$$ We know the cohomology ring of $K(\mathbb{Z},3)$ with coefficients in $\mathbb{Z}/2$ is polynomial ring with generators $Sq^{I}\imath_{3}$ where $\imath_{3}$ is generator of $H^{3}(K(\mathbb{Z},3),\mathbb{Z}/2)$ and $I$ is amissible with excess strictly less than $3$. Using Bockstein homomorphism $\beta = Sq^{1}$ associates with $0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$ we have: $$Sq^{1}Sq^{2}\imath_{3}=Sq^{3}\imath_{3}=\imath_{3}^{2}$$ $$Sq^{1}(\imath_{3}Sq^{2}\imath_{3})=\imath_{3}Sq^{1}Sq^{2}\imath_{3}=\imath_{3}^{3}$$ $$Sq^{1}Sq^{4}Sq^{2}\imath_{3}=Sq^{5}Sq^{2}\imath_{3}=(Sq^{2}\imath_{3})^{2}$$ $\textbf{First Question}$:Look at dimension of those above he concludes that $\text{Ker}\beta = \text{Im}\beta$ through dimension $5 \to 9$ and $2$-torsion in these dimensions consist of elements of order $2$?

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$\textbf{Second question}$: The open circles in the diagram (page $E_{6}$) are $\mathbb{Z}$ cohomology but have been reduced to $\mathbb{Z}/2$ classes, the image of coefficient homomorphism $\mathbb{Z} \to \mathbb{Z}/2$. Why this induced map is injective on $\mathbb{Z}/2$ summands and the image is same as image of Bockstein homomorphism $\beta = Sq^{1}$?

$\textbf{Last question}:$ To finish, we look at the differential $d_{6}^{0,5}$ and conclude if it is not an isomorphism then something survives to $E_{\infty}$ and nonzero torsion appears in $H^{5}(X_{4})$ or $H^{6}(X_{4})$. Why $H^{6}$ of $X_{4}$ here? I thought it must be $H^{6}$ of $K(\mathbb{Z},3)$?

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    $\begingroup$ Is your first 'question' actually a question? $\endgroup$ – Tyrone Nov 25 '18 at 12:25
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If I recall, Hatcher caclulates the integral cohomlogy of $K(\mathbb{Z},3)$ up to degrees 11 or so at some point earlier in the notes. He also tells you at some point that the mod 2 cohomology ring of this space is a free commutative algebra on classes $sq^I\iota$ for admissible multi-indexes $I=(i_1,\dots,i_n)$ satisfying certain 'excess' conditions. In particular, up to degree $10$ we have

$$H^*(K(\mathbb{Z},3);\mathbb{Z}_2)=\mathbb{Z}_2\{\iota,Sq^2\iota,\iota^2, \iota\cdot Sq^2\iota,\iota^3,Sq^4Sq^2\iota,(Sq^2\iota)^2\dots\}$$

where $|\iota|=3$. Here $\iota_3\in H^3(K(\mathbb{Z},3))$ is the fundamental class, and I have written $\iota=\rho_2\iota_3\in H^3(\mathbb{Z},3);\mathbb{Z}_2)$ for its mod 2 reduction. Anyway, $|Sq^2\iota|=5$, $|\iota^2|=6$, $|\iota\cdot Sq^2\iota|=8$, $|\iota^3|=9$, $|Sq^4Sq^2\iota|=9$ and $|(Sq^2\iota)^2|=10$. Now using the Adem relations together with the ring structure we get

\begin{align} Sq^1\iota&=0 \\ Sq^1(Sq^2\iota)&=(Sq^1Sq^2)\iota=Sq^3\iota=\iota^2\\ Sq^1\iota^2&=2(Sq^1\iota)\iota=0\\ Sq^1(\iota\cdot Sq^2\iota)&=(Sq^1\iota)(Sq^2\iota)+\iota(Sq^1Sq^2\iota)=\iota(Sq^3\iota)=\iota^3\\ Sq^1\iota^3&=3(sq^1\iota)\iota=0\\ Sq^1(Sq^4Sq^2\iota)&=Sq^5(Sq^2\iota)=(Sq^2\iota)^2. \end{align}

Thus in these dimensions $\ker(Sq^1)$ is generated as a $\mathbb{Z}_2$-vector space by $\{\iota,\iota^2,\iota^3\}$, whist $im(Sq^1)$ is generated by $\{Sq^1(Sq^2\iota)=\iota^2,Sq^1(\iota\cdot Sq^2\iota)=\iota^3\}$. We see that these agree in dimension $5-9$.

Now recall Hatcher's discussion of the Bockstein spectral sequence, which takes as input $E^*_1=H^*(K(\mathbb{Z},3);\mathbb{Z}_2)$ and converges to $E^*_\infty= H^*(K(\mathbb{Z},3))/(torsion)\otimes\mathbb{Z}_2$. The differential on the $E_1$-page of this spectral sequence is the Bockstein $\beta$, which, in the case $p=2$ is exactly the Steenrod operators $Sq^1$. Moreover, the $E_r$-page can be identified with the subgroup $2^{r-1}\cdot H^*(K(\mathbb{Z},3);\mathbb{Z}_{2^r})$ of $H^*(K(\mathbb{Z},3);\mathbb{Z}_{2^r})$. The previous calculations tell us that $E_2^*=0$ for $*=5,\dots,9$. Therefore $2\cdot H^*(K(\mathbb{Z},3);\mathbb{Z}_4)=0$ for $*=5,\dots, 9$ so the 2-torsion of order at most $2$.

I'm not sure I quite understand your second question. If you can be a bit clearer I will be happy to fill in any further details later on. I think you need to consider this. The exact coefficient sequence $0\rightarrow\mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}\rightarrow\mathbb{Z}_2\rightarrow 0$ induces a long exact sequence for any space $X$

$$\dots\rightarrow H^n(X)\xrightarrow{\times 2}H^n(X)\xrightarrow{\rho_2} H^n(X;\mathbb{Z}_2)\xrightarrow{\delta} H^{n+1}(X)\rightarrow\dots$$

where $\rho_2$ is the mod 2 reduction and $\delta$ is the connecting map of the long exact sequence. Then the Bockstein $\beta=Sq^1$ is the composite

$$\beta=Sq^1=\rho_2\circ\delta:H^n(X;\mathbb{Z}_2)\xrightarrow{\delta} H^{n+1}(X)\xrightarrow{\rho_2} H^{n+1}(X;\mathbb{Z}_2)$$

The point is that not only does it hold that $Sq^1Sq^1=(\rho_2\delta)(\rho_2\delta)=\rho_2(\delta\rho_2)\delta=\rho_2(0)\delta=0$, but also that $Sq^1\rho_2=(\rho_2\delta)\rho_2=(0)\rho_2=0$ and $\delta Sq^1=\delta(\rho_2\delta)=(\delta\rho_2)\delta=(0)\delta=0$. Using what you know about the kernel and image of $Sq^1$ you should be able to use these equations to decide why certain mod 2 reductions are injective.

Third question: You are studying the Serre Spectral sequences of the fibration $K(\mathbb{Z}_2,4)\rightarrow X_4\rightarrow K(\mathbb{Z},3)$, where $X_4$ is obtained as the pullback of the path-space fibration of a map $k_3=Sq^2\iota:K(\mathbb{Z},3)\rightarrow K(\mathbb{Z}_2,5)$. This spectral sequence takes as input $E_2^{p,q}=H^p(K(\mathbb{Z},3);H^q(K(\mathbb{Z}_2,4))$ and converges to $E_\infty^{p,q}=H^{p+q}(X_4)$. This is reason that you end up with classes in $H^6(X_4)$ rather than $H^6(K(\mathbb{Z},3)$. It is the cohomology of the space $X_4$ that you are calculating with this spectral sequence.

With regards to the converges in low degrees of this spectral sequence, there are non non-trivial differentials until the $E_5$-page. The class $\iota\in E_4^{3,0}\cong E_2^{3,0}\cong H^3(K\mathbb{Z},3))$ clearly survives to $E_\infty$. This gives a class in $H^3(K(\mathbb{Z},3;\mathbb{Z}_2)$, and the edge homomorphisms tell you that this is the class $p^*\iota_3$, where $p:X_4\rightarrow K(\mathbb{Z},3)$ is the projection.

Moving onto the $E_5$-page, since $d_5(\iota_4)=Sq^2\iota_3$ by the construction of the fibration, we have $E_\infty^{0,4}\cong E_6^{0,4}\cong \ker(d_5)=0$, and since there are no other classes in $E_5$ with total degree $4$,we find that $H^4(X_4;\mathbb{Z}_2)\cong \oplus_{p+q=4}E_\infty^{p,q}=0$. Here we are on the $E_5$-page.

Now we must work on the $E_6$-page. In total degree $5$ there is only one class, namely, $Sq^1\iota_4\in E_6^{0,5}\cong H^5(K(\mathbb{Z}_2,4);\mathbb{Z}_2)$. The class $Sq^2\iota_3\in E_5^{5,0}\cong H^5(K(\mathbb{Z},3);\mathbb{Z}_2)$ was killed on the $E_5$-page so does no appear at $E_6$.

Now you should recall that the differentials of the spectral sequence commute with the action of the Steenrod algebra. This gives us $d_6(Sq^1\iota_4)=Sq^1(d_6\iota_4)=Sq^1(Sq^2\iota_3)=\iota_3^2$. Therefore $E_\infty^{0,5}\cong E_7^{0,5}\cong\ker(d_6)=0$. Since no other classes of total degree 5 survive to $E_\infty$ we have $H^5(X_4;\mathbb{Z}_2)=0$.

In total degree $6$ we must move to the $E_7$-page. Tthere is again only a single class, $Sq^2\iota_4\in E_7^{0,6}\cong H^6(K(\mathbb{Z}_2,4);\mathbb{Z}_2)$, since $\iota^3\in E_6^{6,0}\cong H^6(K(\mathbb{Z},3),\mathbb{Z}_2)$ was killed on the previous page. We have $d_6(Sq^2\iota_4)=0$ since $E_6^{6,0}=0$.

Thus we see that $H^6(X_4;\mathbb{Z}_2)\cong E_\infty^{6,0}\cong\mathbb{Z}_2$, and is generated by a class $x_6$ satisfying $i^*x_6=Sq^2\iota_4$, where $i:K(\mathbb{Z}_2,4)\rightarrow X_4$ is the fibre inclusion. This class $x_6$ turns out to be the class that must be killed at the next stage of the Postnikov tower.

Summing up we have $H^4(X_4;\mathbb{Z}_2)=0=H^5(X_4;\mathbb{Z}_2)$. The long exact Bockstein sequence tells us that if it is non-trivial then $H^4(X_4)$ must be odd torsion. However we know it cannot be odd torsion, so it must be trivial. Similarly for $H^5(X_4)$. Thus $H^4(X_4)= 0= H^5(X_4)$ (integral coefficients).

Now to sort out what is happening in $H^6(X_4)$ we look at the spectral sequence again. We have $d_8(Sq^3\iota_4)=Sq^3Sq^2\iota_3=Sq^4Sq^1\iota_3=0$. Hence $Sq^3\iota_4$ survives to $E_\infty$ and represents a class $x_7$ satisfying $i^*x_7=Sq^3\iota_4$. Now we have $i^*(Sq^1x_6)=Sq^1(Sq^2\iota_4)=Sq^3\iota_4=i^*x_7$, so that $Sq^1x_6\neq 0$.

In particular, since $Sq^1=\rho_2\delta$, the class $x_6$ cannot be the mod 2 reduction of an integral class - or in fact any $\mathbb{Z}_{2^r}$, $r\geq 2$, class. Therefore $H^6(X_6)=0$ and so $x_6\in H^6(X_6;\mathbb{Z}_2)$ is indeed the class to kill to move to the next Postnikov stage.

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  • $\begingroup$ $\text{Im}Sq^{1}$ is generated by $\left \{ \imath^{2},\imath^{3} \right \} \neq \left \{ \imath,\imath^{2},\imath^{3} \right \}$ and I haven't read Bockstein spectral sequence ( in Hatcher's note is also not yet written ) so I'm so thankful if you could interpret in another way. Lastly, maybe some latex mistakes between $X_{6}$ and $X_{4}$ ? $\endgroup$ – bangbang1412 Nov 25 '18 at 14:17
  • $\begingroup$ @DavidGeal, I apolgise for the typo. $im(Sq^1)$ is generated by the classes I have written. The class $\iota$ is of degree $3$, and the statement is about the classes in degrees 5 to 9. Like I have written, these are $Sq^1(Sq^2\iota)=\iota^2$ and $Sq^1(\iota\cdot Sq^2\iota)=\iota^3$. $\endgroup$ – Tyrone Nov 25 '18 at 15:09
  • $\begingroup$ What I have already written about the Bockstein SS should be enough to help you understand Hatcher's calculations a little better. I shall try to write a little more. $\endgroup$ – Tyrone Nov 25 '18 at 15:14
  • $\begingroup$ @DavidGeal, I've added some details to the bottom of my answer with regards to your last question. $\endgroup$ – Tyrone Nov 25 '18 at 16:16
  • $\begingroup$ it's even true that $H_{i}(X_{4},\mathbb{Z}) \cong H_{i}(S^{3},\mathbb{Z}) \forall i \leq 5$ since $X_{n}$ is obtained as $S^{3}$ after attached cells of dimension greater than $n+2$. Thanks for your answer, you're very enthusiastic. $\endgroup$ – bangbang1412 Nov 25 '18 at 16:29

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