Understanding suspension isomorphism We know that $n$-th ordinary cohomology group $H^{n}(X,G)$ has a representation $[X,K(G,n)]$ and then $H^{n}(X,G) = [X,K(G,n)] = [\Sigma X,K(G,n+1)] = H^{n+1}(\Sigma X,G)$. Besides that, there is an isomorphism $H^{n}(X) \to H^{n+1}(\Sigma X)$ via cross product with a generator of $H^{1}(S^{1})$. I wonder whether two isomorphisms above coincide? 
 A: The two isomorphisms do coincide. To see so, it suffices to understand the universal case, that is for $X=K(G,n)$. Assume that $G$ is a finitely generated abelian group. Then we use the classification theorem for such objects, and the fact that for abelian groups $A$, $B$ there is a homotopy equivalence $K(A\oplus B,n)\simeq K(A,n)\times K(B,n)$, to reduce to the case that $G$ is cyclic on one generator. Thus for convenience we have $G=\mathbb{Z}$ or $G=\mathbb{Z}_{p^k}$ in the following.
We begin with some general observations. For a space $X$ let $\epsilon_X:\Sigma\Omega X\rightarrow X$ be the evaluation map $t\wedge \omega\mapsto\omega(t)$. This map is the adjoint of the identity on $\Omega X$. Regarding this map, G.W. Whitehead has produced a useful homotopy pullback square of the form
$\require{AMScd}$
\begin{CD}
    \Sigma\Omega X@>>> X\vee X\\
    @V\epsilon_X V V   @VV j_X V\\
    X @>\Delta_X>> X\times X
\end{CD}
where $\Delta_X$ is the diagonal and $j_X$ is the natural inclusion. (Recall that a homotopy pullback square result by turning one the maps, say $\Delta_X$, into a fibration. It's an enlightening exercise to work through the details for this case.)
Observe then that the fact that this square is a homotopy pullback tells us that the connectivity of the map $\epsilon_X$ is the same as that of $j_X$, which, if $X$ is $(n-1)$-connected, is $2n-1$. You can use homology, say, to verify this last fact.
The point is that if we take $X=K(G,n+1)$ then it is $n$-connected, and the evaluation map $\epsilon_{n+1}=\epsilon_{K(G,n+1)}$ is $(2n+1)$-connected and so induces isomorphisms
$$H^r(K(G,n);G)\cong H^r(\Sigma\Omega K(G,n+1);G)\cong H^{r-1}(\Omega K(G,n+1);G)$$
for $r<2n+1$. In particular, if $\iota_{n+1}\in H^{n+1}(K(G,n+1);G)$ is the fundamental class, then $\epsilon_{n+1}^*\iota_{n+1}$ is a generator of $H^{n+1}(\Sigma\Omega K(G,n+1);G)\cong G$.
Now choose a homotopy equivalence $\theta:K(G,n)\xrightarrow{\simeq}\Omega K(G,n+1)$ and consider its adjoint $\theta^\#:\Sigma K(G,n)\rightarrow K(G,n+1)$. Observe that
$$\theta^{\#}=\epsilon_{n+1}\circ \Sigma \theta:\Sigma K(G,n)\rightarrow \Sigma\Omega K(G,n+1)\rightarrow K(G,n+1),$$
and that this map induces an isomorphism
$(\theta^{\#})^*:H^{n+1}(K(G,n+1);G)\xrightarrow{\epsilon_{n+1}^*}H^{n+1}(\Sigma\Omega K(G,n+1);G)\xrightarrow{\Sigma \theta^*} H^{n+1}(\Sigma K(G,n);G).$
In general for a space $X$ let us write
$$\Sigma :H^n(X;G)\xrightarrow{\cong} H^{n+1}(\Sigma X;G),\qquad x\mapsto s\wedge x$$
for the suspension isomorphism induced by smashing with the generator $s\in H^1(S^1;G)$. In the case of interest this is $\Sigma :H^n(K(G,n);G)\cong H^{n+1}(\Sigma K(G,n);G)$, $\iota_n\mapsto s\wedge \iota_n$. Thus given the previous isomorphism, the classes $(\theta^{\#})^*\iota_{n+1}$ and $\Sigma\iota_n=s\wedge \iota_n$ differ only by multiplication by a unit in $G$. For our purposes we can redefine the map $\theta$, composing it by the map induced by multiplication by this unit, to get another homotopy equivalence with the desired properties. That is, we can assume without loss of generality that
$$(\theta^{\#})^*\iota_{n+1}=s\wedge \iota_n=\Sigma \iota_n.$$
Now the point is that it is the map $\theta$ which induces the "other" suspension isomorphism. Namely for a space $X$ the isomorphism
$$\sigma:H^n(X;G)\cong [X,K(G,n)]\xrightarrow{\theta_*}[X,\Omega K(G,n+1)]\cong[\Sigma X,K(G,n+1)]\cong H^{n+1}(\Sigma X;G)$$
which sends $f:X\rightarrow K(G,n)$ to the adjoint $(\theta\circ f)^{\#}:\Sigma X\rightarrow K(G,n+1)$. Observe, however, that
$$(\theta\circ f)^{\#}=\theta^\#\circ\Sigma f:\Sigma X\rightarrow\Sigma K(G,n)\rightarrow K(G,n+1)$$
so that if $x\in H^n(X;G)$ is represented by $f$ as above, in that $x=f^*\iota_n$, then $\sigma x\in H^{n+1}(\Sigma X;G)$ is represented by $((\theta\circ f)^{\#})^*\iota_{n+1}=(\theta^\#\circ\Sigma f)^*\iota_{n+1}=\Sigma f^*(\theta^\#)^*\iota_{n+1}$. But we have already seen how $(\theta^\#)^*$ acts. In fact we clearly see that
$\sigma x=((\theta\circ f)^{\#})^*\iota_{n+1}=\Sigma f^*(\theta^\#)^*\iota_{n+1}=\Sigma f^*(s\wedge \iota_n)=s\wedge f^*\iota_n=\Sigma( f^*\iota_n)=\Sigma x$
and conclude that the two suspension isomorphisms $\Sigma$ and $\sigma$ are identical.
