cohomology of pullback

Question

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 $K(\Bbb Z,2)$?

Answer 1

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 $K(\mathbb{Z},2)$.

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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 $K(\mathbb{Z},2)$.