Computing induced map on homology In Hatcher's chapter on spectral sequences, he says that the Serre spectral sequence satisfies naturality properties and later uses it to calculate the induced map on homology. As a specific example, consider the fibration $p:K(\mathbb{Z},2)\to K(\mathbb{Z},2)$ inducing multiplication by $2$ on $\pi_2$. The fiber is $K(\mathbb{Z_2},1)$. Every differential from rows $\ge 1$ must be trivial and every differential from a $\mathbb{Z}$ in the $0^{th}$ row to a $\mathbb{Z_2}$ in an upper row must be nontrivial. Thus $E^{\infty}_{2n,0}$ is the subgroup of $E^2_{2n,0}$ of index $2^n$ and hence concludes about the image of $p_*$.
I cannot figure out how $E^{\infty}_{2n,0}$ is the subgroup of $E^2_{2n,0}$ of index $2^n$. Is the nontrivial map from the $0^{th}$ multiplication by 2. I am guessing the group is $\mathbb{Z}/2^n\mathbb{Z}$. This is a link to the chapter on spectral sequences. The example is on page 538.
 A: No, the subgroup $E^\infty_{2n, 0}$ is $2^n \mathbb{Z} \subseteq \mathbb{Z} = E^2_{2n, 0}$.
The idea is that none of the $\mathbb{Z}/2$'s on the $E^2$ page can survive, since the homology of $K(\mathbb{Z}, 2)$ is torsion-free.  So they must all be hit be differentials, and the only possible source are the $\mathbb{Z}$'s on the $0$-th row.  The differentials $\mathbb{Z} \to \mathbb{Z}/2$ send a generator to a generator, e.g. $1 \mapsto 1 \pmod{2}$.
When this happens, the target $\mathbb{Z}/2$ is killed and the source $\mathbb{Z}$ becomes $\ker(\mathbb{Z} \to \mathbb{Z}/2)$, which is the index two subgroup $2\mathbb{Z} \subseteq \mathbb{Z}$.  Note that this is abstractly isomorphic to $\mathbb{Z}$.  On the next page, we have nontrivial another differential $2\mathbb{Z} \cong \mathbb{Z} \to \mathbb{Z}/2$, so now this kernel is an index two subgroup of $2\mathbb{Z}$, which as a subgroup of $E^2_{2n,0}$ is $4\mathbb{Z} \subseteq \mathbb{Z}$.  This continues to happen as we turn the pages until all the $\mathbb{Z}/2$'s are killed.
From the layout of the spectral sequence, we see that the $\mathbb{Z}$ originating on $E^2_{2n,0}$ has to kill $n$ copies of $\mathbb{Z}/2$.  Therefore, what is left is the index $2^n$ subgroup $2^n \mathbb{Z} \subseteq \mathbb{Z}$.
Of course, all this is completely expected, since the spectral sequence converges to $H_* K(\mathbb{Z}, 2)$ which we already know.  But we learn here that the edge homomorphism induced by $p$, $$H_{2n} K(\mathbb{Z}, 2) \cong E^2_{2n, 0} \to E^\infty_{2n, 0} \xrightarrow{\cong} H_{2n} K(\mathbb{Z}, 2)$$ has image $2^n \mathbb{Z}$, so the map is multiplication by $2^n$.  As Hatcher remarks, this calculation is more easily deduced from the cup product in cohomology.
