Htykuut

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 $2$?

$textbf{Second question}$: 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 $K(mathbb{Z},3)$?

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 $sq^Iiota$ for admissible multi-indexes $I=(i_1,dots,i_n)$ satisfying certain 'excess' conditions. In particular, up to degree $10$ we have

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

where $|iota|=3$. Here $iota_3in H^3(K(mathbb{Z},3))$ is the fundamental class, and I have written $iota=rho_2iota_3in H^3(mathbb{Z},3);mathbb{Z}_2)$ for its mod 2 reduction. Anyway, $|Sq^2iota|=5$, $|iota^2|=6$, $|iotacdot Sq^2iota|=8$, $|iota^3|=9$, $|Sq^4Sq^2iota|=9$ and $|(Sq^2iota)^2|=10$. Now using the Adem relations together with the ring structure we get

begin{align}
Sq^1iota&=0 \
Sq^1(Sq^2iota)&=(Sq^1Sq^2)iota=Sq^3iota=iota^2\
Sq^1iota^2&=2(Sq^1iota)iota=0\
Sq^1(iotacdot Sq^2iota)&=(Sq^1iota)(Sq^2iota)+iota(Sq^1Sq^2iota)=iota(Sq^3iota)=iota^3\
Sq^1iota^3&=3(sq^1iota)iota=0\
Sq^1(Sq^4Sq^2iota)&=Sq^5(Sq^2iota)=(Sq^2iota)^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 ${Sq^1(Sq^2iota)=iota^2,Sq^1(iotacdot Sq^2iota)=iota^3}$. 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 $E^*_infty= H^*(K(mathbb{Z},3))/(torsion)otimesmathbb{Z}_2$. 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 $2cdot H^*(K(mathbb{Z},3);mathbb{Z}_4)=0$ 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 $0rightarrowmathbb{Z}xrightarrow{times 2}mathbb{Z}rightarrowmathbb{Z}_2rightarrow 0$ induces a long exact sequence for any space $X$

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

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

$$beta=Sq^1=rho_2circdelta:H^n(X;mathbb{Z}_2)xrightarrow{delta} H^{n+1}(X)xrightarrow{rho_2} H^{n+1}(X;mathbb{Z}_2)$$

The point is that not only does it hold that $Sq^1Sq^1=(rho_2delta)(rho_2delta)=rho_2(deltarho_2)delta=rho_2(0)delta=0$, but also that $Sq^1rho_2=(rho_2delta)rho_2=(0)rho_2=0$ and $delta Sq^1=delta(rho_2delta)=(deltarho_2)delta=(0)delta=0$. 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 $K(mathbb{Z}_2,4)rightarrow X_4rightarrow K(mathbb{Z},3)$, where $X_4$ is obtained as the pullback of the path-space fibration of a map $k_3=Sq^2iota:K(mathbb{Z},3)rightarrow K(mathbb{Z}_2,5)$. 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 $iotain E_4^{3,0}cong E_2^{3,0}cong H^3(Kmathbb{Z},3))$ 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 $p:X_4rightarrow K(mathbb{Z},3)$ is the projection.

Moving onto the $E_5$-page, since $d_5(iota_4)=Sq^2iota_3$ 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, $Sq^1iota_4in E_6^{0,5}cong H^5(K(mathbb{Z}_2,4);mathbb{Z}_2)$. The class $Sq^2iota_3in E_5^{5,0}cong H^5(K(mathbb{Z},3);mathbb{Z}_2)$ 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 $d_6(Sq^1iota_4)=Sq^1(d_6iota_4)=Sq^1(Sq^2iota_3)=iota_3^2$. Therefore $E_infty^{0,5}cong E_7^{0,5}congker(d_6)=0$. 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, $Sq^2iota_4in E_7^{0,6}cong H^6(K(mathbb{Z}_2,4);mathbb{Z}_2)$, since $iota^3in E_6^{6,0}cong H^6(K(mathbb{Z},3),mathbb{Z}_2)$ was killed on the previous page. We have $d_6(Sq^2iota_4)=0$ since $E_6^{6,0}=0$.

Thus we see that $H^6(X_4;mathbb{Z}_2)cong E_infty^{6,0}congmathbb{Z}_2$, and is generated by a class $x_6$ satisfying $i^*x_6=Sq^2iota_4$, 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 $d_8(Sq^3iota_4)=Sq^3Sq^2iota_3=Sq^4Sq^1iota_3=0$. Hence $Sq^3iota_4$ survives to $E_infty$ and represents a class $x_7$ satisfying $i^*x_7=Sq^3iota_4$. Now we have $i^*(Sq^1x_6)=Sq^1(Sq^2iota_4)=Sq^3iota_4=i^*x_7$, so that $Sq^1x_6neq 0$.

In particular, since $Sq^1=rho_2delta$, the class $x_6$ cannot be the mod 2 reduction of an integral class - or in fact any $mathbb{Z}_{2^r}$, $rgeq 2$, class. Therefore $H^6(X_6)=0$ and so $x_6in H^6(X_6;mathbb{Z}_2)$ is indeed the class to kill to move to the next Postnikov stage.