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I need to show that if we let $T_n(X,A)$ denote the torsion subgroup of $H_n(X,A)$, then the functor $(X,A)\mapsto T_n(X,A)$ with the obvious induced homomorphisms and boundary maps do not define a homology theory.

Using the axioms that Hatcher gives us, my guess is to find maps

$\tilde{H}_n(X/A)\rightarrow \tilde{H}_{n-1}(A)$

$\downarrow\hspace{3 cm}\downarrow$

$\tilde{H}_n(Y/B)\rightarrow \tilde{H}_{n-1}(B)$.

such that either $\tilde{H}_{n-1}(A)$ has zero torsion or $\tilde{H}_n(Y/B)$ has zero torsion, but not both. Then, the diagram for $T_i$ will not be commutative (I also have to make sure there aren't any trivial maps). The simplest map I can think of would be

$\mathbb{Z}_2\rightarrow \mathbb{Z}$

$\downarrow\hspace{1 cm}\downarrow$

$\mathbb{Z}_2\rightarrow \mathbb{Z}_2$.

but I don't know any spaces that have $H_i=\mathbb{Z}_2$ for $i$ even (at least Hatcher didn't go over any such spaces before this question). I just want to know if this is the right path to take, or if I'm wrong and commutativity does, in fact, hold, and I should consider something else.

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The exact sequence $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}_2\rightarrow 0$ doesn't stay exact upon passing to torsion subgroups. Can you find such an exact sequence using the LES of a pair? –  user641 Apr 12 '11 at 3:17
    
I didn't even think of considering the simpler concept of the long exact sequence, but now I've got it. Thanks. –  user9402 Apr 12 '11 at 5:41
    
I think it is usual practice to add answers to questions that have been "resolved", even those where you did all the resolving! So I will add an answer below. :-) –  user641 Apr 12 '11 at 6:00
    
A comment on your title: the set of non-torsion elements of a group is not necessarily a subgroup, even in the abelian case. –  Grumpy Parsnip Apr 12 '11 at 16:02

1 Answer 1

up vote 2 down vote accepted

For $T_n$ to be a homology theory, it must respect the exactness axiom; that is to any pair of spaces $(X,A)$, there should be an induced long exact sequence $$ \cdots\rightarrow T_n(A)\rightarrow T_n(X)\rightarrow T_n(X,A)\rightarrow T_{n-1}(A)\rightarrow\cdots.$$ But if we take the short exact sequence $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}_2\rightarrow 0$, then passing to torsion subgroups we get the non-exact sequence $0\rightarrow 0\rightarrow 0\rightarrow\mathbb{Z}_2\rightarrow 0$. So to show $T_n$ is not a homology theory, it is enough to exhibit the above exact sequence in the LES for ordinary homology of a pair. So take $X$ to be the Möbius strip, and $A$ to be its boundary; then the LES for ordinary (reduced) homology simplifies to the short exact sequence $$ H_1(A)\rightarrow H_1(X)\rightarrow H_1(X,A)\rightarrow 0.$$ Now just observe that $H_1(A)=\mathbb{Z}$, $H_1(X)=\mathbb{Z}$, and $H_1(X,A)=\mathbb{Z}_2$.

You can show $H_1(X,A)=\mathbb{Z}_2$ in two different ways. First, note that if I take the "central" circle of the Möbius strip $X$, then $X$ deformation retracts to this circle. The boundary $A$ corresponds to wrapping twice about this circle. So the inclusion map $i:\ A\rightarrow X$ induces a map on homology $i_*:\ H_1(A)\rightarrow H_1(X)$ which is multiplication by $2$; so in the above SES, $H_1(X,A)$ is the cokernel of this map, namely $\mathbb{Z}_2$.

Alternatively, $H_1(X,A)$ is the first homology of the space obtained by taking the cone over $A$ (call it $CA$), and attaching the boundary to the copy of $A$ in $X$. Now $A$ is just a circle, so $CA$ is just a regular cone, topologically a disk $D^2$. Attaching this disk to $A$ in $X$ means attaching the boundary of $D^2$ to the boundary of the Möbius strip $X$; this is more commonly known as $\mathbb{RP}^2$. So $H_1(X,A)\cong H_1(\mathbb{RP}^2)=\mathbb{Z}_2$.

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