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.
Best Answer
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.