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?










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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















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?










share|cite|improve this question






















  • 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













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?










share|cite|improve this question













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






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David Geal

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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


















  • 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










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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.






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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.






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      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.






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        up vote
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        up vote
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        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.






        share|cite|improve this answer












        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.







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        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Tyrone

        4,06011125




        4,06011125






























             

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