# Filling the Details #4: Maps of Eilenberg MacLane Spectra induces stable cohomology operations

First of all, sorry for not having written lately. I’ve just done the GRE and GRE math examinations, together with the TOEFL. I really hope the result will be good, since I’m going to apply in the USA!

Ok, back to Math 🙂  As the title suggests, I want to write something about this known result, since I needed it for understanding why the first non-vanishing differential in the Atiyah-Hirzebruch Spectral Sequence is a stable operation. What I’m going to show is not the nicest way to prove it, but still I think it’s worth to take a look at a more “concrete” proof

First of all it’s better to revise some concepts: recall the concept of the fundamental class $\imath_{n,A}$ for a E-M space $K(A,n)$. You can find something  here at the bottom of the article.

Let $\rho_{A,n} \colon \Sigma K(A,n) \to K(A,n+1)$ be the canonical structure map for the E-M spectrum representing $H^*(-,A)$. In other words, $\rho_{A,n} \colon \Sigma K(A,n)\to K(A,n+1)$ is the inclusion. (To convince yourself just use a little bit of cellular approximation and Freudenthal Suspension Theorem)

Now we need to introduce some version of the suspension isomorphism in cohomology.

Suspension Isomorphism 1: Let us denote with

$\Sigma^* \colon \colon \widetilde{H}^n(X) \to \widetilde{H}^{n+1}(\Sigma X)$

the map sending a class $x$ to $1\wedge x$, where $1 \in H^1(S^1)$ is a generator. In Tammo tom Dieck’s Algebraic Topology, page 414, Proposition 17.3.1 it’s proven that $\sigma(x)=1\wedge x$, where $\sigma$ is tom Dieck’s suspension defined at page 249, 10.2.5. Therefore we have $\Sigma^*=\sigma$.

Suspension Isomorphism 2: In Harper’s Secondary Cohomology Operation, at page 22 there is defined the so called Mayer-Vietoris suspension:

$\Delta_* \colon \widetilde{H}^n(X) \to \widetilde{H}^{n+1}(\Sigma X)$

as the following composition:

where $i_1 \colon (T_0X,X)\to (\Sigma X,T_1 X)$ and $i_2\colon (\Sigma X, *) \to (\Sigma X, T_1 X)$ are the inclusions and $T_0 X, T_1 X$ are the lower and upper cone of $X$ respectively.

Claim 1:  Given $f \colon X \to Z$, then $\Delta_* \circ f^* = (\Sigma f)^*\circ \Delta^*$.

Proof: It’s just an application of the naturality of the l.e.s of the pair and of the respective triple, as indicate in this diagram:

where starting at the lower left position and going clockwise gives you $\Delta_* \circ f^*$, going counter-clockwise gives you $latex (\Sigma f)^*\circ \Delta^*$.

Claim 2: $\Delta^*=\Sigma^*$

Proof:  (Idea) It’s enough to show that $\Delta^* = \sigma$. A little bit of diagram chasing and naturality of the boundary operators for the map $X \times I \to T_0X$ which pinch the lower extremity of the cone should do the trick. We make use of the definition of the boundary operator for the l.e.s. of the triple in cohomology  (have a look here for the homology one, (we are looking at the second triangle pointing down in the top row)

We are almost ready to tackle down our original problem. We need the following technical lemmas. Let $\Phi \colon H^n(X,A) \to \hom_Z(H_n(X,A),A)$ be the map of the UCT, and let $H$ the Hurewicz isomorphism from homotopy to homology.

Claim 3: The following diagram commutes for an appropriate choice of the polarising isomorphism:

Proof: The upper triangle commutes since Hurewicz map can be made compatible with the boundary operator and therefore the suspensions (Have a look at Harper’s Secondary Cohomology Operator page 24 in the middle). The lower trapezoid commutes because we can inductively define $\phi_{n+1}$ to do so, since $(\rho_A)_h$ is an isomorphism in degree $n+1$ (use cellular approximation).

Claim 4: We have $\Phi(-)((\Sigma_*)^{-1} -)=\Phi(\Sigma^*-)(-)$

Proof: This is the diagram we want to prove commutativity:

which boils down to show that the two elements in the lower right corner in this diagram are the same:

but this is true since we know that every element $y \in H_{n+1}(\Sigma X; Z)$ is of the form $1 \times x$ for some $x \in H_{n}(X; Z)$. Using it we are done.

Claim 5: With the above notation, we have $\rho_A^*(\imath_{A,n+1})=\Sigma^*\imath_{A,n}$

Proof: Now we start to make use of the results done so far. We have the following equalities:

$\rho_A^*(\imath_{A,n+1})= \rho^*\Phi^{-1}(\phi_{n+1} \circ H^{-1} )$

where we sued the definition of the fundamental class. Now we use naturality of UCT and Hurewicz to get

$\rho_A^*(\imath_{A,n+1})= \Phi^{-1}(\phi_{n+1}\circ (\rho_A)_h \circ H^{-1})$

By Claim 3 we have

$\rho_A^*(\imath_{A,n+1})= \Phi^{-1}(\phi_{n+1}\circ\phi_{n+1}^{-1}\circ \phi_n \circ H^{-1} \circ \Sigma_*^{-1} \circ H \circ H^{-1})$

which is equal to

$\rho_A^*(\imath_{A,n+1})= \Phi^{-1}(\phi_n \circ H^{-1} \circ \Sigma_*^{-1})$

Now observe that, after composing with $\Phi$ both sides, it’s equivalent to prove that

$\phi_n \circ H^{-1} \circ \Sigma_*^{-1} = \Phi( \Sigma^* \circ \Phi^{-1} (\phi_n \circ H^{-1}))$

but using Claim 4 we can move the suspension isomorphism inside the first UCT to the right:

$\Phi( \Sigma^* \circ \Phi^{-1} (\phi_n \circ H^{-1})) = \Phi (\Phi^{-1} )(\phi_n\circ H^{-1}))\Sigma_*^{-1}$

Therefore we have the conclusion.

We now return to the stable cohomology operations stuff (finally).

Claim 6 $\{\theta_n\}_n$ is stable cohomology operation if and only if

$\Sigma^*( \theta_n(\imath_{A,n}))=\rho_A^*(\theta_{n+1}(\imath_{A,n+1}))$

Proof: It’s easy to see that $\{\theta_n\}_n$ is stable cohomology operation if and only if it commutes with the suspension isomorphism. Now we have the following chain of equivalencies:

$\rho^*_A\theta_{n+1}(\imath_{A,n+1})= \theta_{n+1}(\rho_A^*(\imath_{A,n+1}))=\theta_{n+1}\Sigma^*(\imath_{A,n})$

where we used naturality and Claim 5.  This shows that the two conditions in the thesis are equivalent as claimed.

Now we are done for good.

Theorem: Let now $\{\theta_n\}_n$ be a map of spectra between (shifted) E-M spectra, then in cohomology, the components $\theta_n$‘s build together in a stable cohomology operation.

Proof: By definition of map of spectra we have the following commutative diagram:

which gives the following relation:

$\rho_A^*\theta_{n+1}(\imath_{B,m+1})=(\Sigma \theta_n)^*\rho_B (\imath_{B,m+1})$

Now using Claim 5 and Claim 1 we have that

$(\Sigma \theta_n)^{*}\rho_B^* (\imath_{B,m+1}) = (\Sigma \theta_n)^*\Sigma^*(\imath_{B,m}) = \Sigma^*(\theta_n (\imath_{B,m}))$

Since we proved that $\rho_A^*\theta_{n+1}(\imath_{B,m+1})= \Sigma^*(\theta_n (\imath_{B,m}))$ by Claim 6 we conclude that $\{ \theta_n\}_n$ is a stable cohomology operation as claimed

The more theoretical approach would be to prove that the cohomology functor is representable and therefore natural transformations between representable functors are in natural bijection with appropriate maps between the representing objects i.e. spectra.