cohomology of pullback Does the pullback $X$ of the diagram
$$K(\Bbb Z,2) \stackrel{\cdot n}{\longrightarrow} K(\Bbb Z,2) \stackrel{\cdot 2}{\longleftarrow} K(\Bbb Z,2)$$
have the same integral cohomology groups as $K(\Bbb Z,2)$?
 A: The space $X$ has the same integral cohomology groups as $K(\mathbb{Z},2)$ if and only if $n$ is odd. 
The short exact sequence of groups
$$ 0 \to \mathbb{Z} \stackrel{n}{\to} \mathbb{Z} \stackrel{q_n}{\to} \mathbb{Z}/n \to 0 $$
where $q_n$ is the quotient, induces a fibration
$$ K(\mathbb{Z},1) \stackrel{n}{\to} K(\mathbb{Z},1) \stackrel{q_n}{\to} K(\mathbb{Z}/n,1) $$
which in turn gives a fibration
$$ K(\mathbb{Z},2) \stackrel{n}{\to} K(\mathbb{Z},2) \stackrel{Bq_n}{\to} K(\mathbb{Z}/n,2) $$
In particular, the homotopy fibre of $K(\mathbb{Z},2) \stackrel{2}{\to} K(\mathbb{Z},2)$ is a $K(\mathbb{Z}/2,1)$ and this fibration is a pullback on the universal fibration: 
$$
\require{AMScd}
\begin{CD}
K(\mathbb{Z}/2,1) @>>> K(\mathbb{Z}/2,1) \\
@VVV @VVV \\
K(\mathbb{Z},2) @>>> \Omega K(\mathbb{Z}/2,2) \\
@VV2V @VVV \\
K(\mathbb{Z},2) @>Bq_2>> K(\mathbb{Z}/2,2) \\
\end{CD}
$$
Therefore the pullback that you are seeking is the pullback of the universal fibration under the composition $Bq_2 \circ n$. This composition is induced from the homomorphism
$$ \mathbb{Z} \to \mathbb{Z}/2 $$
$$ a \mapsto q_2(na) $$
Note that this map is $q_2$ if $n$ is odd and the trivial homomorphism if $n$ is even. Therefore if $n$ is odd, the pullback under $Bq_2 \circ n$ is homotopically equivalent to the pullback under $Bq_2$, which is $K(\mathbb{Z},2)$.
If $n$ is even, the pullback under $Bq_2 \circ n$ is homotopically equivalent to the pullback under the constant map, which is $K(\mathbb{Z},2) \times K(\mathbb{Z}/2,1) = \mathbb{C}P^{\infty} \times \mathbb{R}P^{\infty}$.
Using Kunneth's theorem for instance, the second cohomology group of this space is $\mathbb{Z} \oplus \mathbb{Z}/2$, so it does not have the cohomology groups of $K(\mathbb{Z},2)$.

Here is an alternative solution in case it may help anyone. When you have a principal bundle $q \colon P \to B$, the pullback of the diagram
$$ 
\require{AMScd}
\begin{CD}
 & &  P \\
@. @VqVV \\
P @>q>> B
\end{CD}
$$
is the trivial principal bundle over $P$ because it has a section. 
The map $K(\mathbb{Z},2) \stackrel{2}{\to} K(\mathbb{Z},2)$ is a principal bundle with fibre $K(\mathbb{Z}/2,1)$, hence when $n=2$, we have $X = K(\mathbb{Z},2) \times K(\mathbb{Z}/2,1)$.
When $n$ is even, say $n=2m$, the map $K(\mathbb{Z},2) \stackrel{n}{\to} K(\mathbb{Z},2)$ is the composition 
$$K(\mathbb{Z},2) \stackrel{m}{\to} K(\mathbb{Z},2) \stackrel{2}{\to} K(\mathbb{Z},2)$$ 
So doing the pullback under the map $n$ is the same as doing the pullback under the map $2$, which is trivial, and then under the map $m$, therefore also trivial. So in this case we also have $X = K(\mathbb{Z},2) \times K(\mathbb{Z}/2,2)$.
Now when $n$ is odd we use the long exact sequence of homotopy groups for the fibration
$$ K(\mathbb{Z}/2,1) \to X \to K(\mathbb{Z},2) $$
from where we get that $X$ is path-connected and $\pi_k(X)=0$ for $k \geq 3$. For the remaining two homotopy groups we use the naturality of the long exact sequence with respect to maps of fibrations for:
$$ \require{AMScd}
\begin{CD}
K(\mathbb{Z}/2,1) @>>> K(\mathbb{Z}/2,1) \\
@VVV @VVV \\
X @>>> K(\mathbb{Z},2) \\
@VVV @V2VV \\
K(\mathbb{Z},2) @>n>> K(\mathbb{Z},2) \\
\end{CD}
$$
obtaining a commutative diagram with exact rows
$$ \require{AMScd}
\begin{CD}
0 @>>> \pi_2(X) @>>> \mathbb{Z} @>>> \mathbb{Z}/2 @>>> \pi_1(X) @>>> 0 \\
@. @VVV @VnVV @VidVV @VVV \\
0 @>>> \mathbb{Z} @>2>> \mathbb{Z} @>q>> \mathbb{Z}/2 @>>> 0 @>>> 0
\end{CD}
$$
where $q \colon \mathbb{Z} \to \mathbb{Z}/2$ is the quotient homomorphism. Then the homomorphism $\mathbb{Z} \to \mathbb{Z}/2$ in the upper row is forced to be  $ k \mapsto q(nk)=q(k)$. From this we get that $\pi_1(X)=0$ and $\pi_2(X) \cong \mathbb{Z}$, that is, $X$ is a $K(\mathbb{Z},2)$.
