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given a topological space $X$, $H_n(X,-)$ is a functor from the category of abelian groups to itself. i want to clarify the following :

1) given an homomorphism $f:G\rightarrow H$ of abelian groups what is explicitly the induced map $f_*:H_n(X,G)\rightarrow H_n(X,H)$

my guess: an element in $H_n(X,G)$ is a formal sum $\sum{g_i c_i}$ where $g_i\in G$ and $c_i$ a class of a cyle in $C_n(X,G)$ the free $\mathbb Z$-module on $n$-singular simplices. so $f_*(\sum{g_i c_i})=\sum{f(g_i) c_i}$

2)given an abelian group $A$, what is the canonical homomorphism $f:\mathbb Z \rightarrow A$ that is used to induce $f_*:H_n(X,\mathbb Z)\rightarrow H_n(X,A)$ and then induce $$f_{**}:H_n(X,\mathbb Z)\otimes A\rightarrow H_n(X,A)$$ that gives the short exact sequence in the universal coefficient theorem in homology:

$$0 \rightarrow H_i(X, \mathbb{Z})\otimes A\rightarrow H_i(X,A)\rightarrow\mbox{Tor}(H_{i-1}(X, \mathbb{Z}),A)\rightarrow 0$$

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If you want to claim something, then 1) is not it: it is not a claim but a question! –  Mariano Suárez-Alvarez Jun 8 '11 at 14:54
    
@Mariano: palio wrote "clarify", not "claim". –  Rasmus Jun 8 '11 at 15:25
    
@Mariano : Hi mariano! that indeed was a question :) –  palio Jun 8 '11 at 16:50

2 Answers 2

up vote 4 down vote accepted

1) Yes, your guess is correct.

2) There is no non-zero canonical morphism $\mathbb Z\to A$.

The map $\phi:H_\bullet(X,\mathbb Z)\otimes A\to H_\bullet(X,A)$ is constructed differently. Let $\alpha\otimes a$ be an elementary tensor in the domain. The homology class $\alpha$ is the class of some cycle $\sum_ic_i\sigma_i$ with $c_i\in\mathbb Z$. Then $\phi(\alpha\otimes a)$ is the class in $H_\bullet(X,A)$ of the element $\sum_ic_ia\sigma_i$. You can easily check that this is in fact a cycle in the complex which computes $H_\bullet(X,A)$, so this makes sense.

 

The very best to understand what are the maps in the Universal Coefficient Theorem is to follow a proof of the theorem in detail: all proofs I know of actually construct the maps!

 

Later: I claimed there is no non-zero canonical map $\mathbb Z\to A$. Let me prove at least there is no natural map (for there is no sensible definition of canonical!). Suppose for all abelian groups $\phi_A:\mathbb Z\to A$ is a group homomomorphism which depends naturally on $A$. Let $A$ be a group, and let $i_1,i_2:A\to A\oplus A$ be the two obvious inclusions into the "coordinate axes". Then naturality implies that $$i_1(\phi_A(1))=\phi_{A\oplus A}(1)=i_2(\phi_A(1)),$$ so $\phi_{A\oplus A}(1)\in i_1(A)\cap i_2(A)=0$. It follows that $\phi_{A\oplus A}(1)=0$ and, since $i_1(\phi_A(1))=\phi_{A\oplus A}(1)$ and $i_1$ is injective, that $\phi_A(1)=0$. Thus $\phi_A=0$.

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The first point is essentially correct. Let $C_n(X)$ be the $n$-chains in $X$ with integral coefficients. For an abelian group $A$, by definition, $C_n(X,A) = C_n(X) \otimes A$. To the chain complex $(C_*(X), \partial)$ there is an associated chain complex $(C_*(X,A), \partial \otimes 1_A)$, where $1_A$ is the identity map on $A$. The homology of the first complex is, by definition, $H_*(X)$ (with integral coefficients), and the homology of the second complex is, by defintion, $H_*(X,A)$.

Given a homomorphism $f:A \to B$ of abelian groups, there is an induced map $f_*: H_*(X,A) \to H_*(X,B)$ given by $f_*(\sum \sigma_i \otimes a_i) = \sum \sigma_i \otimes f(a_i)$. Here the notation is a little sloppy: I am regarding $\tau = \sum \sigma_i \otimes a_i$ as an element of $C_n(X,A)$, not as a homology class. So, you need to check that this map is well-defined. (This is a highly recommended exercise.)

As to the second point, the universal coefficient theorem is comparing $H_n(X) \otimes A$ with $H_n(X,A)$. The difference is the order in which we tensored with $A$. The map $H_n(X) \otimes A \to H_n(X,A)$ is... (EDIT: thanks to Mariano's remarks below) given by mapping $[\tau] \otimes a$ to the class $[\tau \otimes a]$, where $\tau$ is a cycle in $C_n(X)$ and $[\tau]$ is the homology class that it represents.

(Here is my original, incorrect statement: "induced by the map on the chain level: $C_n(X) \otimes A \to C_n(X) \otimes A$, i.e. the identity map.")

Again, there is work to be done: you need to check that this induces a well-defined map on homology.

To get a feel for what is going on, it makes sense to have an example at hand. Comparing $H_2(\mathbb{R}P^2)$ to $H_2(\mathbb{R}P^2, \mathbb{Z}_2)$ should be helpful.

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In (2), it is quite not true that the map is induced by the identity. –  Mariano Suárez-Alvarez Jun 8 '11 at 15:14
    
Doesn't the identity induce the same map as you write in your answer below? The cycle $(\sum c_i \sigma_i) \otimes a$ is mapped to $\sum (c_i \sigma_i \otimes a)$. What am I missing? –  Robert Bell Jun 8 '11 at 15:24
    
@Robert: The thing is, in the domain of the map, you first take the homology of the complex $C_\bullet(X)$ and then apply to the result the functor $(\mathord-)\otimes A$, while in the codomain of the map you first apply the functor $(\mathord-)\otimes A$ and then take the homology. While morally the map is "motivated" by the identity, at the moment you want to construct it there is no identity to speak of, as the domain and codomain are rather different objects. –  Mariano Suárez-Alvarez Jun 8 '11 at 15:33
    
Agreed. Thank you, that clears things up for me. I'll edit the answer accordingly. –  Robert Bell Jun 8 '11 at 15:40
    
thanks guys!! by my question i was hoping to understand the following statement : There exists a map $X\rightarrow K(H_i(X,\mathbb Q),i)$ corresponding by the univ coeff theorem to $H_i(X,\mathbb Z)\rightarrow H_i(X,\mathbb Q)$ induced from the inclusion $\mathbb Z\rightarrow \mathbb Q$ but this is still eluding me! –  palio Jun 8 '11 at 16:45

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