Mixing
Understanding suspension isomorphism
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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 isomorphism above coincide?
algebraic-topology homology-cohomology
asked yesterday
David Geal
314
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up vote
2
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favorite
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 isomorphism above coincide?
algebraic-topology homology-cohomology
asked yesterday
David Geal
314
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?
– freakish
yesterday
@freakish See math.stackexchange.com/q/45556.
– Paul Frost
yesterday
For $n=0$ you need reduced homology.
– Paul Frost
yesterday
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 isomorphism above coincide?
algebraic-topology homology-cohomology
asked yesterday
David Geal
314
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 isomorphism above coincide?
algebraic-topology homology-cohomology
algebraic-topology homology-cohomology
asked yesterday
David Geal
314
asked yesterday
David Geal
314
asked yesterday
David Geal
314
asked yesterday
David Geal
314
asked yesterday
David Geal
314
314
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?
– freakish
yesterday
@freakish See math.stackexchange.com/q/45556.
– Paul Frost
yesterday
For $n=0$ you need reduced homology.
– Paul Frost
yesterday
add a comment |
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?
– freakish
yesterday
@freakish See math.stackexchange.com/q/45556.
– Paul Frost
yesterday
For $n=0$ you need reduced homology.
– Paul Frost
yesterday
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?
– freakish
yesterday
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?
– freakish
yesterday
@freakish See math.stackexchange.com/q/45556.
– Paul Frost
yesterday
@freakish See math.stackexchange.com/q/45556.
– Paul Frost
yesterday
For $n=0$ you need reduced homology.
– Paul Frost
yesterday
For $n=0$ you need reduced homology.
– Paul Frost
yesterday
add a comment |
1 Answer
1
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oldest
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up vote
0
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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 $K(Aoplus B,n)simeq K(A,n)times K(B,n)$, 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 $epsilon_X:SigmaOmega Xrightarrow X$ be the evaluation map $twedge omegamapstoomega(t)$. 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}
SigmaOmega X@>>> Xvee X\
@Vepsilon_X V V @VV j_X V\
X @>Delta_X>> Xtimes 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
$$H^r(K(G,n);G)cong H^r(SigmaOmega K(G,n+1);G)cong H^{r-1}(Omega K(G,n+1);G)$$
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 $H^{n+1}(SigmaOmega K(G,n+1);G)cong G$.
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
$$theta^{#}=epsilon_{n+1}circ Sigma theta:Sigma K(G,n)rightarrow SigmaOmega K(G,n+1)rightarrow K(G,n+1),$$
and that this map induces an isomorphism
$(theta^{#})^*:H^{n+1}(K(G,n+1);G)xrightarrow{epsilon_{n+1}^*}H^{n+1}(SigmaOmega K(G,n+1);G)xrightarrow{Sigma theta^*} H^{n+1}(Sigma K(G,n);G).$
In general for a space $X$ let us write
$$Sigma :H^n(X;G)xrightarrow{cong} H^{n+1}(Sigma X;G),qquad xmapsto swedge x$$
for the suspension isomorphism induced by smashing with the generator $sin H^1(S^1;G)$. In the case of interest this is $Sigma :H^n(K(G,n);G)cong H^{n+1}(Sigma K(G,n);G)$, $iota_nmapsto swedge iota_n$. Thus given the previous isomorphism, the classes $(theta^{#})^*iota_{n+1}$ and $Sigmaiota_n=swedge iota_n$ 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
$$(theta^{#})^*iota_{n+1}=swedge iota_n=Sigma iota_n.$$
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 $f:Xrightarrow K(G,n)$ to the adjoint $(thetacirc f)^{#}:Sigma Xrightarrow K(G,n+1)$. Observe, however, that
$$(thetacirc f)^{#}=theta^#circSigma f:Sigma XrightarrowSigma K(G,n)rightarrow K(G,n+1)$$
so that if $xin H^n(X;G)$ is represented by $f$ as above, in that $x=f^*iota_n$, then $sigma xin H^{n+1}(Sigma X;G)$ is represented by $((thetacirc f)^{#})^*iota_{n+1}=(theta^#circSigma f)^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}$. But we have already seen how $(theta^#)^*$ acts. In fact we clearly see that
$sigma x=((thetacirc f)^{#})^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}=Sigma f^*(swedge iota_n)=swedge f^*iota_n=Sigma( f^*iota_n)=Sigma x$
and conclude that the two suspension isomorphisms $Sigma$ and $sigma$ are identical.
answered 1 hour ago
Tyrone
4,06011125
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1 Answer
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1 Answer
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active
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up vote
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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 $K(Aoplus B,n)simeq K(A,n)times K(B,n)$, 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 $epsilon_X:SigmaOmega Xrightarrow X$ be the evaluation map $twedge omegamapstoomega(t)$. 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}
SigmaOmega X@>>> Xvee X\
@Vepsilon_X V V @VV j_X V\
X @>Delta_X>> Xtimes 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
$$H^r(K(G,n);G)cong H^r(SigmaOmega K(G,n+1);G)cong H^{r-1}(Omega K(G,n+1);G)$$
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 $H^{n+1}(SigmaOmega K(G,n+1);G)cong G$.
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
$$theta^{#}=epsilon_{n+1}circ Sigma theta:Sigma K(G,n)rightarrow SigmaOmega K(G,n+1)rightarrow K(G,n+1),$$
and that this map induces an isomorphism
$(theta^{#})^*:H^{n+1}(K(G,n+1);G)xrightarrow{epsilon_{n+1}^*}H^{n+1}(SigmaOmega K(G,n+1);G)xrightarrow{Sigma theta^*} H^{n+1}(Sigma K(G,n);G).$
In general for a space $X$ let us write
$$Sigma :H^n(X;G)xrightarrow{cong} H^{n+1}(Sigma X;G),qquad xmapsto swedge x$$
for the suspension isomorphism induced by smashing with the generator $sin H^1(S^1;G)$. In the case of interest this is $Sigma :H^n(K(G,n);G)cong H^{n+1}(Sigma K(G,n);G)$, $iota_nmapsto swedge iota_n$. Thus given the previous isomorphism, the classes $(theta^{#})^*iota_{n+1}$ and $Sigmaiota_n=swedge iota_n$ 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
$$(theta^{#})^*iota_{n+1}=swedge iota_n=Sigma iota_n.$$
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 $f:Xrightarrow K(G,n)$ to the adjoint $(thetacirc f)^{#}:Sigma Xrightarrow K(G,n+1)$. Observe, however, that
$$(thetacirc f)^{#}=theta^#circSigma f:Sigma XrightarrowSigma K(G,n)rightarrow K(G,n+1)$$
so that if $xin H^n(X;G)$ is represented by $f$ as above, in that $x=f^*iota_n$, then $sigma xin H^{n+1}(Sigma X;G)$ is represented by $((thetacirc f)^{#})^*iota_{n+1}=(theta^#circSigma f)^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}$. But we have already seen how $(theta^#)^*$ acts. In fact we clearly see that
$sigma x=((thetacirc f)^{#})^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}=Sigma f^*(swedge iota_n)=swedge f^*iota_n=Sigma( f^*iota_n)=Sigma x$
and conclude that the two suspension isomorphisms $Sigma$ and $sigma$ are identical.
answered 1 hour ago
Tyrone
4,06011125
add a comment |
up vote
0
down vote
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 $K(Aoplus B,n)simeq K(A,n)times K(B,n)$, 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 $epsilon_X:SigmaOmega Xrightarrow X$ be the evaluation map $twedge omegamapstoomega(t)$. 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}
SigmaOmega X@>>> Xvee X\
@Vepsilon_X V V @VV j_X V\
X @>Delta_X>> Xtimes 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
$$H^r(K(G,n);G)cong H^r(SigmaOmega K(G,n+1);G)cong H^{r-1}(Omega K(G,n+1);G)$$
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 $H^{n+1}(SigmaOmega K(G,n+1);G)cong G$.
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
$$theta^{#}=epsilon_{n+1}circ Sigma theta:Sigma K(G,n)rightarrow SigmaOmega K(G,n+1)rightarrow K(G,n+1),$$
and that this map induces an isomorphism
$(theta^{#})^*:H^{n+1}(K(G,n+1);G)xrightarrow{epsilon_{n+1}^*}H^{n+1}(SigmaOmega K(G,n+1);G)xrightarrow{Sigma theta^*} H^{n+1}(Sigma K(G,n);G).$
In general for a space $X$ let us write
$$Sigma :H^n(X;G)xrightarrow{cong} H^{n+1}(Sigma X;G),qquad xmapsto swedge x$$
for the suspension isomorphism induced by smashing with the generator $sin H^1(S^1;G)$. In the case of interest this is $Sigma :H^n(K(G,n);G)cong H^{n+1}(Sigma K(G,n);G)$, $iota_nmapsto swedge iota_n$. Thus given the previous isomorphism, the classes $(theta^{#})^*iota_{n+1}$ and $Sigmaiota_n=swedge iota_n$ 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
$$(theta^{#})^*iota_{n+1}=swedge iota_n=Sigma iota_n.$$
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 $f:Xrightarrow K(G,n)$ to the adjoint $(thetacirc f)^{#}:Sigma Xrightarrow K(G,n+1)$. Observe, however, that
$$(thetacirc f)^{#}=theta^#circSigma f:Sigma XrightarrowSigma K(G,n)rightarrow K(G,n+1)$$
so that if $xin H^n(X;G)$ is represented by $f$ as above, in that $x=f^*iota_n$, then $sigma xin H^{n+1}(Sigma X;G)$ is represented by $((thetacirc f)^{#})^*iota_{n+1}=(theta^#circSigma f)^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}$. But we have already seen how $(theta^#)^*$ acts. In fact we clearly see that
$sigma x=((thetacirc f)^{#})^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}=Sigma f^*(swedge iota_n)=swedge f^*iota_n=Sigma( f^*iota_n)=Sigma x$
and conclude that the two suspension isomorphisms $Sigma$ and $sigma$ are identical.
answered 1 hour ago
Tyrone
4,06011125
add a comment |
up vote
0
down vote
up vote
0
down vote
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 $K(Aoplus B,n)simeq K(A,n)times K(B,n)$, 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 $epsilon_X:SigmaOmega Xrightarrow X$ be the evaluation map $twedge omegamapstoomega(t)$. 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}
SigmaOmega X@>>> Xvee X\
@Vepsilon_X V V @VV j_X V\
X @>Delta_X>> Xtimes 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
$$H^r(K(G,n);G)cong H^r(SigmaOmega K(G,n+1);G)cong H^{r-1}(Omega K(G,n+1);G)$$
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 $H^{n+1}(SigmaOmega K(G,n+1);G)cong G$.
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
$$theta^{#}=epsilon_{n+1}circ Sigma theta:Sigma K(G,n)rightarrow SigmaOmega K(G,n+1)rightarrow K(G,n+1),$$
and that this map induces an isomorphism
$(theta^{#})^*:H^{n+1}(K(G,n+1);G)xrightarrow{epsilon_{n+1}^*}H^{n+1}(SigmaOmega K(G,n+1);G)xrightarrow{Sigma theta^*} H^{n+1}(Sigma K(G,n);G).$
In general for a space $X$ let us write
$$Sigma :H^n(X;G)xrightarrow{cong} H^{n+1}(Sigma X;G),qquad xmapsto swedge x$$
for the suspension isomorphism induced by smashing with the generator $sin H^1(S^1;G)$. In the case of interest this is $Sigma :H^n(K(G,n);G)cong H^{n+1}(Sigma K(G,n);G)$, $iota_nmapsto swedge iota_n$. Thus given the previous isomorphism, the classes $(theta^{#})^*iota_{n+1}$ and $Sigmaiota_n=swedge iota_n$ 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
$$(theta^{#})^*iota_{n+1}=swedge iota_n=Sigma iota_n.$$
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 $f:Xrightarrow K(G,n)$ to the adjoint $(thetacirc f)^{#}:Sigma Xrightarrow K(G,n+1)$. Observe, however, that
$$(thetacirc f)^{#}=theta^#circSigma f:Sigma XrightarrowSigma K(G,n)rightarrow K(G,n+1)$$
so that if $xin H^n(X;G)$ is represented by $f$ as above, in that $x=f^*iota_n$, then $sigma xin H^{n+1}(Sigma X;G)$ is represented by $((thetacirc f)^{#})^*iota_{n+1}=(theta^#circSigma f)^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}$. But we have already seen how $(theta^#)^*$ acts. In fact we clearly see that
$sigma x=((thetacirc f)^{#})^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}=Sigma f^*(swedge iota_n)=swedge f^*iota_n=Sigma( f^*iota_n)=Sigma x$
and conclude that the two suspension isomorphisms $Sigma$ and $sigma$ are identical.
answered 1 hour ago
Tyrone
4,06011125
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 $K(Aoplus B,n)simeq K(A,n)times K(B,n)$, 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 $epsilon_X:SigmaOmega Xrightarrow X$ be the evaluation map $twedge omegamapstoomega(t)$. 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}
SigmaOmega X@>>> Xvee X\
@Vepsilon_X V V @VV j_X V\
X @>Delta_X>> Xtimes 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
$$H^r(K(G,n);G)cong H^r(SigmaOmega K(G,n+1);G)cong H^{r-1}(Omega K(G,n+1);G)$$
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 $H^{n+1}(SigmaOmega K(G,n+1);G)cong G$.
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
$$theta^{#}=epsilon_{n+1}circ Sigma theta:Sigma K(G,n)rightarrow SigmaOmega K(G,n+1)rightarrow K(G,n+1),$$
and that this map induces an isomorphism
$(theta^{#})^*:H^{n+1}(K(G,n+1);G)xrightarrow{epsilon_{n+1}^*}H^{n+1}(SigmaOmega K(G,n+1);G)xrightarrow{Sigma theta^*} H^{n+1}(Sigma K(G,n);G).$
In general for a space $X$ let us write
$$Sigma :H^n(X;G)xrightarrow{cong} H^{n+1}(Sigma X;G),qquad xmapsto swedge x$$
for the suspension isomorphism induced by smashing with the generator $sin H^1(S^1;G)$. In the case of interest this is $Sigma :H^n(K(G,n);G)cong H^{n+1}(Sigma K(G,n);G)$, $iota_nmapsto swedge iota_n$. Thus given the previous isomorphism, the classes $(theta^{#})^*iota_{n+1}$ and $Sigmaiota_n=swedge iota_n$ 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
$$(theta^{#})^*iota_{n+1}=swedge iota_n=Sigma iota_n.$$
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 $f:Xrightarrow K(G,n)$ to the adjoint $(thetacirc f)^{#}:Sigma Xrightarrow K(G,n+1)$. Observe, however, that
$$(thetacirc f)^{#}=theta^#circSigma f:Sigma XrightarrowSigma K(G,n)rightarrow K(G,n+1)$$
so that if $xin H^n(X;G)$ is represented by $f$ as above, in that $x=f^*iota_n$, then $sigma xin H^{n+1}(Sigma X;G)$ is represented by $((thetacirc f)^{#})^*iota_{n+1}=(theta^#circSigma f)^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}$. But we have already seen how $(theta^#)^*$ acts. In fact we clearly see that
$sigma x=((thetacirc f)^{#})^*iota_{n+1}=Sigma f^*(theta^#)^*iota_{n+1}=Sigma f^*(swedge iota_n)=swedge f^*iota_n=Sigma( f^*iota_n)=Sigma x$
and conclude that the two suspension isomorphisms $Sigma$ and $sigma$ are identical.
answered 1 hour ago
Tyrone
4,06011125
answered 1 hour ago
Tyrone
4,06011125
answered 1 hour ago
Tyrone
4,06011125
answered 1 hour ago
Tyrone
4,06011125
4,06011125
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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?
– freakish
yesterday
@freakish See math.stackexchange.com/q/45556.
– Paul Frost
yesterday
For $n=0$ you need reduced homology.
– Paul Frost
yesterday