What does “splitting naturally” mean in the Universal Coefficients Theorem

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The Universal Coefficients Theorem states that



$0rightarrow H_n(X)otimes Grightarrow H_n(X;G)rightarrowoperatornameTor(H_n-1(X),G)rightarrow 0$



splits, but not naturally. In all the algebraic topology contexts I've come across, "natural" implies commutativity, but I don't see what a "natural split" means. Can someone give me an explicit definition?



Edit:



From what I've read... if I understand this correctly, splitting naturally implies that if



$0rightarrow Arightarrow Aoplus Crightarrow Crightarrow 0$



and



$0rightarrow A'rightarrow A'oplus C'rightarrow C'rightarrow 0$



and given maps $a:Arightarrow A'$ and $c:Crightarrow C'$, the map $Aoplus Crightarrow A'oplus C'$ has to be the map $(a,c)$ in order for the diagram to commute.










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  • 1




    Here's a relevant discussion on math.MO
    – t.b.
    Apr 18 '11 at 19:11










  • I already did see that discussion, but I still don't understand how the OP's question relates to the Universal Coefficients Theorem. I don't see how those induced maps on homology and homology with coefficients relate to the exact sequence in the UCT. (Also, I don't know what $Sigma$ means, so that may help me.)
    – user9402
    Apr 18 '11 at 19:17











  • Ok, so here's the thing. The sequence splits, that is to say that the middle term is the direct sum of the two outer terms. If this splitting were natural in $X$, the map in the middle would have to be the direct sum of the outer maps for all maps $f: X to Y$. Dylan asks about a map that induces the identity on the two outer terms in the UCT, but fails to be the identity in the middle, so this shows that the splitting isn't natural in $X$.
    – t.b.
    Apr 18 '11 at 19:26











  • The $Sigma$ means the suspension, denoted $S$ on wikipedia.
    – t.b.
    Apr 18 '11 at 19:28






  • 2




    Yes, what you say is correct. However, it becomes clearer if you write $0 to A(X) to B(X) to C(X) to 0$ and $0 to A'(X) to B'(X) to C'(X) to 0$ (the sequences depend naturally on $X$). Now they split. That is to say, $B(X) cong A(X) oplus C(X)$, for some isomorphism, but this isomorphism does not depend naturally on $X$.
    – t.b.
    Apr 18 '11 at 19:53















up vote
11
down vote

favorite
6












The Universal Coefficients Theorem states that



$0rightarrow H_n(X)otimes Grightarrow H_n(X;G)rightarrowoperatornameTor(H_n-1(X),G)rightarrow 0$



splits, but not naturally. In all the algebraic topology contexts I've come across, "natural" implies commutativity, but I don't see what a "natural split" means. Can someone give me an explicit definition?



Edit:



From what I've read... if I understand this correctly, splitting naturally implies that if



$0rightarrow Arightarrow Aoplus Crightarrow Crightarrow 0$



and



$0rightarrow A'rightarrow A'oplus C'rightarrow C'rightarrow 0$



and given maps $a:Arightarrow A'$ and $c:Crightarrow C'$, the map $Aoplus Crightarrow A'oplus C'$ has to be the map $(a,c)$ in order for the diagram to commute.










share|cite|improve this question



















  • 1




    Here's a relevant discussion on math.MO
    – t.b.
    Apr 18 '11 at 19:11










  • I already did see that discussion, but I still don't understand how the OP's question relates to the Universal Coefficients Theorem. I don't see how those induced maps on homology and homology with coefficients relate to the exact sequence in the UCT. (Also, I don't know what $Sigma$ means, so that may help me.)
    – user9402
    Apr 18 '11 at 19:17











  • Ok, so here's the thing. The sequence splits, that is to say that the middle term is the direct sum of the two outer terms. If this splitting were natural in $X$, the map in the middle would have to be the direct sum of the outer maps for all maps $f: X to Y$. Dylan asks about a map that induces the identity on the two outer terms in the UCT, but fails to be the identity in the middle, so this shows that the splitting isn't natural in $X$.
    – t.b.
    Apr 18 '11 at 19:26











  • The $Sigma$ means the suspension, denoted $S$ on wikipedia.
    – t.b.
    Apr 18 '11 at 19:28






  • 2




    Yes, what you say is correct. However, it becomes clearer if you write $0 to A(X) to B(X) to C(X) to 0$ and $0 to A'(X) to B'(X) to C'(X) to 0$ (the sequences depend naturally on $X$). Now they split. That is to say, $B(X) cong A(X) oplus C(X)$, for some isomorphism, but this isomorphism does not depend naturally on $X$.
    – t.b.
    Apr 18 '11 at 19:53













up vote
11
down vote

favorite
6









up vote
11
down vote

favorite
6






6





The Universal Coefficients Theorem states that



$0rightarrow H_n(X)otimes Grightarrow H_n(X;G)rightarrowoperatornameTor(H_n-1(X),G)rightarrow 0$



splits, but not naturally. In all the algebraic topology contexts I've come across, "natural" implies commutativity, but I don't see what a "natural split" means. Can someone give me an explicit definition?



Edit:



From what I've read... if I understand this correctly, splitting naturally implies that if



$0rightarrow Arightarrow Aoplus Crightarrow Crightarrow 0$



and



$0rightarrow A'rightarrow A'oplus C'rightarrow C'rightarrow 0$



and given maps $a:Arightarrow A'$ and $c:Crightarrow C'$, the map $Aoplus Crightarrow A'oplus C'$ has to be the map $(a,c)$ in order for the diagram to commute.










share|cite|improve this question















The Universal Coefficients Theorem states that



$0rightarrow H_n(X)otimes Grightarrow H_n(X;G)rightarrowoperatornameTor(H_n-1(X),G)rightarrow 0$



splits, but not naturally. In all the algebraic topology contexts I've come across, "natural" implies commutativity, but I don't see what a "natural split" means. Can someone give me an explicit definition?



Edit:



From what I've read... if I understand this correctly, splitting naturally implies that if



$0rightarrow Arightarrow Aoplus Crightarrow Crightarrow 0$



and



$0rightarrow A'rightarrow A'oplus C'rightarrow C'rightarrow 0$



and given maps $a:Arightarrow A'$ and $c:Crightarrow C'$, the map $Aoplus Crightarrow A'oplus C'$ has to be the map $(a,c)$ in order for the diagram to commute.







algebraic-topology






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share|cite|improve this question













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edited Apr 18 '11 at 19:48

























asked Apr 18 '11 at 19:06







user9402














  • 1




    Here's a relevant discussion on math.MO
    – t.b.
    Apr 18 '11 at 19:11










  • I already did see that discussion, but I still don't understand how the OP's question relates to the Universal Coefficients Theorem. I don't see how those induced maps on homology and homology with coefficients relate to the exact sequence in the UCT. (Also, I don't know what $Sigma$ means, so that may help me.)
    – user9402
    Apr 18 '11 at 19:17











  • Ok, so here's the thing. The sequence splits, that is to say that the middle term is the direct sum of the two outer terms. If this splitting were natural in $X$, the map in the middle would have to be the direct sum of the outer maps for all maps $f: X to Y$. Dylan asks about a map that induces the identity on the two outer terms in the UCT, but fails to be the identity in the middle, so this shows that the splitting isn't natural in $X$.
    – t.b.
    Apr 18 '11 at 19:26











  • The $Sigma$ means the suspension, denoted $S$ on wikipedia.
    – t.b.
    Apr 18 '11 at 19:28






  • 2




    Yes, what you say is correct. However, it becomes clearer if you write $0 to A(X) to B(X) to C(X) to 0$ and $0 to A'(X) to B'(X) to C'(X) to 0$ (the sequences depend naturally on $X$). Now they split. That is to say, $B(X) cong A(X) oplus C(X)$, for some isomorphism, but this isomorphism does not depend naturally on $X$.
    – t.b.
    Apr 18 '11 at 19:53













  • 1




    Here's a relevant discussion on math.MO
    – t.b.
    Apr 18 '11 at 19:11










  • I already did see that discussion, but I still don't understand how the OP's question relates to the Universal Coefficients Theorem. I don't see how those induced maps on homology and homology with coefficients relate to the exact sequence in the UCT. (Also, I don't know what $Sigma$ means, so that may help me.)
    – user9402
    Apr 18 '11 at 19:17











  • Ok, so here's the thing. The sequence splits, that is to say that the middle term is the direct sum of the two outer terms. If this splitting were natural in $X$, the map in the middle would have to be the direct sum of the outer maps for all maps $f: X to Y$. Dylan asks about a map that induces the identity on the two outer terms in the UCT, but fails to be the identity in the middle, so this shows that the splitting isn't natural in $X$.
    – t.b.
    Apr 18 '11 at 19:26











  • The $Sigma$ means the suspension, denoted $S$ on wikipedia.
    – t.b.
    Apr 18 '11 at 19:28






  • 2




    Yes, what you say is correct. However, it becomes clearer if you write $0 to A(X) to B(X) to C(X) to 0$ and $0 to A'(X) to B'(X) to C'(X) to 0$ (the sequences depend naturally on $X$). Now they split. That is to say, $B(X) cong A(X) oplus C(X)$, for some isomorphism, but this isomorphism does not depend naturally on $X$.
    – t.b.
    Apr 18 '11 at 19:53








1




1




Here's a relevant discussion on math.MO
– t.b.
Apr 18 '11 at 19:11




Here's a relevant discussion on math.MO
– t.b.
Apr 18 '11 at 19:11












I already did see that discussion, but I still don't understand how the OP's question relates to the Universal Coefficients Theorem. I don't see how those induced maps on homology and homology with coefficients relate to the exact sequence in the UCT. (Also, I don't know what $Sigma$ means, so that may help me.)
– user9402
Apr 18 '11 at 19:17





I already did see that discussion, but I still don't understand how the OP's question relates to the Universal Coefficients Theorem. I don't see how those induced maps on homology and homology with coefficients relate to the exact sequence in the UCT. (Also, I don't know what $Sigma$ means, so that may help me.)
– user9402
Apr 18 '11 at 19:17













Ok, so here's the thing. The sequence splits, that is to say that the middle term is the direct sum of the two outer terms. If this splitting were natural in $X$, the map in the middle would have to be the direct sum of the outer maps for all maps $f: X to Y$. Dylan asks about a map that induces the identity on the two outer terms in the UCT, but fails to be the identity in the middle, so this shows that the splitting isn't natural in $X$.
– t.b.
Apr 18 '11 at 19:26





Ok, so here's the thing. The sequence splits, that is to say that the middle term is the direct sum of the two outer terms. If this splitting were natural in $X$, the map in the middle would have to be the direct sum of the outer maps for all maps $f: X to Y$. Dylan asks about a map that induces the identity on the two outer terms in the UCT, but fails to be the identity in the middle, so this shows that the splitting isn't natural in $X$.
– t.b.
Apr 18 '11 at 19:26













The $Sigma$ means the suspension, denoted $S$ on wikipedia.
– t.b.
Apr 18 '11 at 19:28




The $Sigma$ means the suspension, denoted $S$ on wikipedia.
– t.b.
Apr 18 '11 at 19:28




2




2




Yes, what you say is correct. However, it becomes clearer if you write $0 to A(X) to B(X) to C(X) to 0$ and $0 to A'(X) to B'(X) to C'(X) to 0$ (the sequences depend naturally on $X$). Now they split. That is to say, $B(X) cong A(X) oplus C(X)$, for some isomorphism, but this isomorphism does not depend naturally on $X$.
– t.b.
Apr 18 '11 at 19:53





Yes, what you say is correct. However, it becomes clearer if you write $0 to A(X) to B(X) to C(X) to 0$ and $0 to A'(X) to B'(X) to C'(X) to 0$ (the sequences depend naturally on $X$). Now they split. That is to say, $B(X) cong A(X) oplus C(X)$, for some isomorphism, but this isomorphism does not depend naturally on $X$.
– t.b.
Apr 18 '11 at 19:53











2 Answers
2






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oldest

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up vote
2
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$requireAMScd newcommandRPmathbbRP$
Note: The splitting is actually natural in $G$ contrary to what is written in many texts. What she is referring to is the following:$newcommandzmathbbZ$. Lets fix $G=z/2$.



Thm: There is no possibility of a natural transformation from the functor $H_*(-,z/2)$ to $(H_*(-,z))^* oplus Ext(H_*-1(-,z),z/2)$.



Proof: Suppose if possible that for all spaces $X,Y$ and every continuous map $X to Y$, there is an induced a commuting square



$ beginCD
H_*(X,z/2) @>cong>> Tor(H_*-1(X),z/2) oplus H_*(X) otimes z/2 \
@VVV @VVV\
H_*(Y,z/2) @>cong>> Tor(H_*-1(Y),z/2) oplus H_*(Y) otimes z/2 \
endCD$
.



Then there would have to be a commuting square induced by $RP^2 to RP^2/RP^1 to S^2$:



$ beginCD
H_2(RP^2,z/2) @>cong>> Tor(H_1(RP^2),z/2) oplus H_2(RP^2) otimes z/2 \
@| @V0VV\
H_2(S^2,z/2) @>cong>> Tor(H_2-1(S^2),z/2) oplus H_2(S^2) otimes z/2 \
endCD$. Contradiction since the 0 map would need to be an isomorphism.



Note: One could not have expected there to be such a natural transformation because $H_*(X)$ splits into objects of different homogenous degree, while the induced maps of topological maps map elements of the same degree onto elements of the same degree.



Note 2: The diagram $$ beginCD
H^2(RP^2,z/2) @>cong>> Ext(H_1(RP^2),z/2) oplus H_2(RP^2)^* \
@| @V0VV\
H^2(S^2,z/2) @>cong>> Ext(H_1(S^2),z/2) oplus H_2(S^2)^* \
endCD$$ shows that the exact sequence of the cohomological universal coefficient theorem does not split naturally with respect to the topological space.






share|cite|improve this answer





























    up vote
    1
    down vote













    It doesn't seem like anyone has said it the following way, yet, which might help clear things up:



    What does it mean for an exact sequence to split? It means, among other things, that you have given a section of the last map in the sequence. Now, we have a functor from spaces to short exact sequences and we might ask whether or not we can lift this to a functor from spaces to short exact sequences together with a section (i.e. split short exact sequences.)



    However, we cannot. The reason is that there are endomorphisms of spaces that induce the identity map on $H_*(X)$ but not on $H_*(X;G)$. This means, in particular, that the functor would give an endomorphism of exact sequences that is the identity on the ends but not in the middle. It is easy to check that no such endomorphism of a split exact sequence exists, and so the functor does not lift.



    So perhaps a better thing to say is that the sequence does not split "functorially."






    share|cite|improve this answer
















    • 1




      Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’.
      – Zhen Lin
      Mar 9 '13 at 0:32






    • 1




      @ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_n-1(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation.
      – Dylan Wilson
      Mar 9 '13 at 0:56










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    2 Answers
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    up vote
    2
    down vote













    $requireAMScd newcommandRPmathbbRP$
    Note: The splitting is actually natural in $G$ contrary to what is written in many texts. What she is referring to is the following:$newcommandzmathbbZ$. Lets fix $G=z/2$.



    Thm: There is no possibility of a natural transformation from the functor $H_*(-,z/2)$ to $(H_*(-,z))^* oplus Ext(H_*-1(-,z),z/2)$.



    Proof: Suppose if possible that for all spaces $X,Y$ and every continuous map $X to Y$, there is an induced a commuting square



    $ beginCD
    H_*(X,z/2) @>cong>> Tor(H_*-1(X),z/2) oplus H_*(X) otimes z/2 \
    @VVV @VVV\
    H_*(Y,z/2) @>cong>> Tor(H_*-1(Y),z/2) oplus H_*(Y) otimes z/2 \
    endCD$
    .



    Then there would have to be a commuting square induced by $RP^2 to RP^2/RP^1 to S^2$:



    $ beginCD
    H_2(RP^2,z/2) @>cong>> Tor(H_1(RP^2),z/2) oplus H_2(RP^2) otimes z/2 \
    @| @V0VV\
    H_2(S^2,z/2) @>cong>> Tor(H_2-1(S^2),z/2) oplus H_2(S^2) otimes z/2 \
    endCD$. Contradiction since the 0 map would need to be an isomorphism.



    Note: One could not have expected there to be such a natural transformation because $H_*(X)$ splits into objects of different homogenous degree, while the induced maps of topological maps map elements of the same degree onto elements of the same degree.



    Note 2: The diagram $$ beginCD
    H^2(RP^2,z/2) @>cong>> Ext(H_1(RP^2),z/2) oplus H_2(RP^2)^* \
    @| @V0VV\
    H^2(S^2,z/2) @>cong>> Ext(H_1(S^2),z/2) oplus H_2(S^2)^* \
    endCD$$ shows that the exact sequence of the cohomological universal coefficient theorem does not split naturally with respect to the topological space.






    share|cite|improve this answer


























      up vote
      2
      down vote













      $requireAMScd newcommandRPmathbbRP$
      Note: The splitting is actually natural in $G$ contrary to what is written in many texts. What she is referring to is the following:$newcommandzmathbbZ$. Lets fix $G=z/2$.



      Thm: There is no possibility of a natural transformation from the functor $H_*(-,z/2)$ to $(H_*(-,z))^* oplus Ext(H_*-1(-,z),z/2)$.



      Proof: Suppose if possible that for all spaces $X,Y$ and every continuous map $X to Y$, there is an induced a commuting square



      $ beginCD
      H_*(X,z/2) @>cong>> Tor(H_*-1(X),z/2) oplus H_*(X) otimes z/2 \
      @VVV @VVV\
      H_*(Y,z/2) @>cong>> Tor(H_*-1(Y),z/2) oplus H_*(Y) otimes z/2 \
      endCD$
      .



      Then there would have to be a commuting square induced by $RP^2 to RP^2/RP^1 to S^2$:



      $ beginCD
      H_2(RP^2,z/2) @>cong>> Tor(H_1(RP^2),z/2) oplus H_2(RP^2) otimes z/2 \
      @| @V0VV\
      H_2(S^2,z/2) @>cong>> Tor(H_2-1(S^2),z/2) oplus H_2(S^2) otimes z/2 \
      endCD$. Contradiction since the 0 map would need to be an isomorphism.



      Note: One could not have expected there to be such a natural transformation because $H_*(X)$ splits into objects of different homogenous degree, while the induced maps of topological maps map elements of the same degree onto elements of the same degree.



      Note 2: The diagram $$ beginCD
      H^2(RP^2,z/2) @>cong>> Ext(H_1(RP^2),z/2) oplus H_2(RP^2)^* \
      @| @V0VV\
      H^2(S^2,z/2) @>cong>> Ext(H_1(S^2),z/2) oplus H_2(S^2)^* \
      endCD$$ shows that the exact sequence of the cohomological universal coefficient theorem does not split naturally with respect to the topological space.






      share|cite|improve this answer
























        up vote
        2
        down vote










        up vote
        2
        down vote









        $requireAMScd newcommandRPmathbbRP$
        Note: The splitting is actually natural in $G$ contrary to what is written in many texts. What she is referring to is the following:$newcommandzmathbbZ$. Lets fix $G=z/2$.



        Thm: There is no possibility of a natural transformation from the functor $H_*(-,z/2)$ to $(H_*(-,z))^* oplus Ext(H_*-1(-,z),z/2)$.



        Proof: Suppose if possible that for all spaces $X,Y$ and every continuous map $X to Y$, there is an induced a commuting square



        $ beginCD
        H_*(X,z/2) @>cong>> Tor(H_*-1(X),z/2) oplus H_*(X) otimes z/2 \
        @VVV @VVV\
        H_*(Y,z/2) @>cong>> Tor(H_*-1(Y),z/2) oplus H_*(Y) otimes z/2 \
        endCD$
        .



        Then there would have to be a commuting square induced by $RP^2 to RP^2/RP^1 to S^2$:



        $ beginCD
        H_2(RP^2,z/2) @>cong>> Tor(H_1(RP^2),z/2) oplus H_2(RP^2) otimes z/2 \
        @| @V0VV\
        H_2(S^2,z/2) @>cong>> Tor(H_2-1(S^2),z/2) oplus H_2(S^2) otimes z/2 \
        endCD$. Contradiction since the 0 map would need to be an isomorphism.



        Note: One could not have expected there to be such a natural transformation because $H_*(X)$ splits into objects of different homogenous degree, while the induced maps of topological maps map elements of the same degree onto elements of the same degree.



        Note 2: The diagram $$ beginCD
        H^2(RP^2,z/2) @>cong>> Ext(H_1(RP^2),z/2) oplus H_2(RP^2)^* \
        @| @V0VV\
        H^2(S^2,z/2) @>cong>> Ext(H_1(S^2),z/2) oplus H_2(S^2)^* \
        endCD$$ shows that the exact sequence of the cohomological universal coefficient theorem does not split naturally with respect to the topological space.






        share|cite|improve this answer














        $requireAMScd newcommandRPmathbbRP$
        Note: The splitting is actually natural in $G$ contrary to what is written in many texts. What she is referring to is the following:$newcommandzmathbbZ$. Lets fix $G=z/2$.



        Thm: There is no possibility of a natural transformation from the functor $H_*(-,z/2)$ to $(H_*(-,z))^* oplus Ext(H_*-1(-,z),z/2)$.



        Proof: Suppose if possible that for all spaces $X,Y$ and every continuous map $X to Y$, there is an induced a commuting square



        $ beginCD
        H_*(X,z/2) @>cong>> Tor(H_*-1(X),z/2) oplus H_*(X) otimes z/2 \
        @VVV @VVV\
        H_*(Y,z/2) @>cong>> Tor(H_*-1(Y),z/2) oplus H_*(Y) otimes z/2 \
        endCD$
        .



        Then there would have to be a commuting square induced by $RP^2 to RP^2/RP^1 to S^2$:



        $ beginCD
        H_2(RP^2,z/2) @>cong>> Tor(H_1(RP^2),z/2) oplus H_2(RP^2) otimes z/2 \
        @| @V0VV\
        H_2(S^2,z/2) @>cong>> Tor(H_2-1(S^2),z/2) oplus H_2(S^2) otimes z/2 \
        endCD$. Contradiction since the 0 map would need to be an isomorphism.



        Note: One could not have expected there to be such a natural transformation because $H_*(X)$ splits into objects of different homogenous degree, while the induced maps of topological maps map elements of the same degree onto elements of the same degree.



        Note 2: The diagram $$ beginCD
        H^2(RP^2,z/2) @>cong>> Ext(H_1(RP^2),z/2) oplus H_2(RP^2)^* \
        @| @V0VV\
        H^2(S^2,z/2) @>cong>> Ext(H_1(S^2),z/2) oplus H_2(S^2)^* \
        endCD$$ shows that the exact sequence of the cohomological universal coefficient theorem does not split naturally with respect to the topological space.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 9 at 15:07









        fermesomme

        5,10222251




        5,10222251










        answered Jun 28 '16 at 22:47









        Hari Rau-Murthy

        741317




        741317




















            up vote
            1
            down vote













            It doesn't seem like anyone has said it the following way, yet, which might help clear things up:



            What does it mean for an exact sequence to split? It means, among other things, that you have given a section of the last map in the sequence. Now, we have a functor from spaces to short exact sequences and we might ask whether or not we can lift this to a functor from spaces to short exact sequences together with a section (i.e. split short exact sequences.)



            However, we cannot. The reason is that there are endomorphisms of spaces that induce the identity map on $H_*(X)$ but not on $H_*(X;G)$. This means, in particular, that the functor would give an endomorphism of exact sequences that is the identity on the ends but not in the middle. It is easy to check that no such endomorphism of a split exact sequence exists, and so the functor does not lift.



            So perhaps a better thing to say is that the sequence does not split "functorially."






            share|cite|improve this answer
















            • 1




              Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’.
              – Zhen Lin
              Mar 9 '13 at 0:32






            • 1




              @ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_n-1(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation.
              – Dylan Wilson
              Mar 9 '13 at 0:56














            up vote
            1
            down vote













            It doesn't seem like anyone has said it the following way, yet, which might help clear things up:



            What does it mean for an exact sequence to split? It means, among other things, that you have given a section of the last map in the sequence. Now, we have a functor from spaces to short exact sequences and we might ask whether or not we can lift this to a functor from spaces to short exact sequences together with a section (i.e. split short exact sequences.)



            However, we cannot. The reason is that there are endomorphisms of spaces that induce the identity map on $H_*(X)$ but not on $H_*(X;G)$. This means, in particular, that the functor would give an endomorphism of exact sequences that is the identity on the ends but not in the middle. It is easy to check that no such endomorphism of a split exact sequence exists, and so the functor does not lift.



            So perhaps a better thing to say is that the sequence does not split "functorially."






            share|cite|improve this answer
















            • 1




              Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’.
              – Zhen Lin
              Mar 9 '13 at 0:32






            • 1




              @ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_n-1(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation.
              – Dylan Wilson
              Mar 9 '13 at 0:56












            up vote
            1
            down vote










            up vote
            1
            down vote









            It doesn't seem like anyone has said it the following way, yet, which might help clear things up:



            What does it mean for an exact sequence to split? It means, among other things, that you have given a section of the last map in the sequence. Now, we have a functor from spaces to short exact sequences and we might ask whether or not we can lift this to a functor from spaces to short exact sequences together with a section (i.e. split short exact sequences.)



            However, we cannot. The reason is that there are endomorphisms of spaces that induce the identity map on $H_*(X)$ but not on $H_*(X;G)$. This means, in particular, that the functor would give an endomorphism of exact sequences that is the identity on the ends but not in the middle. It is easy to check that no such endomorphism of a split exact sequence exists, and so the functor does not lift.



            So perhaps a better thing to say is that the sequence does not split "functorially."






            share|cite|improve this answer












            It doesn't seem like anyone has said it the following way, yet, which might help clear things up:



            What does it mean for an exact sequence to split? It means, among other things, that you have given a section of the last map in the sequence. Now, we have a functor from spaces to short exact sequences and we might ask whether or not we can lift this to a functor from spaces to short exact sequences together with a section (i.e. split short exact sequences.)



            However, we cannot. The reason is that there are endomorphisms of spaces that induce the identity map on $H_*(X)$ but not on $H_*(X;G)$. This means, in particular, that the functor would give an endomorphism of exact sequences that is the identity on the ends but not in the middle. It is easy to check that no such endomorphism of a split exact sequence exists, and so the functor does not lift.



            So perhaps a better thing to say is that the sequence does not split "functorially."







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 8 '13 at 23:55









            Dylan Wilson

            4,6461726




            4,6461726







            • 1




              Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’.
              – Zhen Lin
              Mar 9 '13 at 0:32






            • 1




              @ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_n-1(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation.
              – Dylan Wilson
              Mar 9 '13 at 0:56












            • 1




              Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’.
              – Zhen Lin
              Mar 9 '13 at 0:32






            • 1




              @ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_n-1(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation.
              – Dylan Wilson
              Mar 9 '13 at 0:56







            1




            1




            Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’.
            – Zhen Lin
            Mar 9 '13 at 0:32




            Rather, one should remark that the word ‘natural’ here is being used in the same technical sense as in ‘natural transformation’ or ‘natural isomorphism’.
            – Zhen Lin
            Mar 9 '13 at 0:32




            1




            1




            @ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_n-1(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation.
            – Dylan Wilson
            Mar 9 '13 at 0:56




            @ZhenLin: Ah but to say that I would have to reformulate slightly. A natural transformation is some morphism of functors. In this case we have one morphism of functors, namely from $H_n(X;G)$ to $Tor(H_n-1(X), G)$. The claim is that there is no natural transformation the other way such that one of the compositions is the identity natural transformation.
            – Dylan Wilson
            Mar 9 '13 at 0:56

















             

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