# 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?

• Is the first composition even an isomorphism of groups? Because $H^{n}(X,G) = [X,K(G,n)]$ is only a bijection. Is there even a group structure on $[X, K(G,n)]$? Or is there some other reason why this composition is an isomorphism? Commented Nov 19, 2018 at 14:36
• @freakish See math.stackexchange.com/q/45556. Commented Nov 19, 2018 at 15:25
• For $n=0$ you need reduced homology. Commented Nov 19, 2018 at 15:30

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.

• Hey Tyrone, I'm having trouble understanding what "smashing with a generator of $H^1(S^1)$" should mean, and I can't find a reference. For homology such a thing makes sense to me, since you can just compose the cross product map with the quotient map, but I can only think of maps that are naturally described as $H^{n+1}(SX) \rightarrow H^{n}(X)$ when it comes to cohomology. Commented Jun 30, 2019 at 20:59
• @ConnorMalin If $X$, $Y$ are spaces and $x\in H^n(X;A)$, $y\in H^m(Y;B)$ then there is an external product giving you a class $x\wedge y\in H^{n+m}(X\wedge Y;A\otimes B)$. This is just a reduced version of the standard cross product. If you represent $x,y$ by maps $f_x,f_y$ to EM spaces, then the smash is the class $x\wedge y:X\wedge Y\xrightarrow{f_x\wedge f_y}K(A,n)\otimes K(B,m)\rightarrow K(n+m,A\otimes B)$, where the last map is a chosen generator for $H^{n+m}(K(A,n)\wedge K(B,m))$ (check Kunneth). Commented Jul 1, 2019 at 9:10
• If $A=B$ is a ring then the map induced by $A\otimes A\rightarrow A$ internalises the product in the standard way. Commented Jul 1, 2019 at 9:13