How can I show that this diagram is commutative using Mayer-Vietoris sequences?

Question

From Rotman's Algebraic Topology:

Assume that $X = X_1^{\circ} \cup X_2^{\circ}$ and $Y = Y_1^{\circ} \cup Y_2^{\circ}$; assume further that $f : X \rightarrow Y$ is continuous with $f(X_i) \subset Y_i$ for $i=1,2$. Then $f^R_* D = D' f_*$, where:

$f_* : H_n(X) \rightarrow H_n(Y)$, and $f^R_* : H_{n-1}(X_1 \cap X_2) \rightarrow H_{n-1}(Y_1 \cap Y_2) $ is the restriction of $f$ and $D, D'$ are connecting homomorphisms of Mayer-Vietoris sequences.

$D$ is defined as $D = dh^{-1}_*q_*$, where $d$ is a connecting homomorphism of the pair $(X_1, X_1 \cap X_2)$, and $h$ and $q$ are inclusions defined by $h : (X_1, X_1 \cap X_2) \rightarrow (X, X_2)$ and $q : (X, \emptyset) \rightarrow (X, X_2)$.

I started by using the two short exact sequences $0 \rightarrow S_*(X_1 \cap X_2) \xrightarrow i S_*(X_1) \oplus S_*(X_2) \xrightarrow p S_*(X) \rightarrow 0 $ and $0 \rightarrow S_*(Y_1 \cap Y_2) \xrightarrow i S_*(Y_1) \oplus S_*(Y_2) \xrightarrow p S_*(Y) \rightarrow 0 $ that are connected by the induced maps from $f : X \rightarrow Y$. We then have a commutative diagram with exact rows :

$\dots \rightarrow H_n(X_1 \cap X_2) \xrightarrow {i_*} H_n(X_1) \oplus H_n(X_2) \xrightarrow {p_*} H_n(X) \xrightarrow \delta H_{n-1}(X_1 \cap X_2) \rightarrow \dots$

$\dots \rightarrow H_n(Y_1 \cap Y_2) \xrightarrow {i'_*} H_n(Y_1) \oplus H_n(Y_2) \xrightarrow {p'_*} H_n(Y) \xrightarrow \delta' H_{n-1}(Y_1 \cap Y_2) \rightarrow \dots$

connected again with induced maps from $f$, where $\delta$ is a connecting homomorpism.

I can see that $\delta' f = f^R \delta$, but I'm not sure how I'd use this to show $f_*^RD = D' f_*$.

Any hints or suggestions?

Answer 1

We know that $D = dh^{-1}_*q_*$ and $D' = d_*'h'^{-1}_*q'_*$. If I can show that that $f_*$ commutes with each $d_*$, $h_*$ and $q_*$, then we will be done.

$q_*$ is induced by inclusion $(X, \varnothing) \hookrightarrow (X, X_2)$ and $q_*'$ by $(Y, \varnothing) \hookrightarrow (Y, Y_2)$. Since, $f(X_i) \subset Y_i$, $f$ induces map at the relative chain level: $ f_\# : C_*(X, X_2) \to C_*(Y, Y_2) $. $f_\#$ commutes with the maps at the relative chain level $q_\#$ and $q'_\#$, that is $f_\# q_\# = q'_\# f_\#$ (this just follows by definition of $f_\#$). Since, the arrows commute at chain level, they certainly do at homology level.

$h_*$ is the excision map (an isomorphism). So, it commutes with $f_*$.

Lastly, you agreed in the comments $f_*$ commutes with the connecting homomorphisms for the LES of pairs $(X, A)$ and $(Y, B)$. In particular put $A = X_1 \cap X_2$ and $B = Y_1 \cap Y_2$ [Observe that $f(X_1 \cap X_2) \subset Y_1 \cap Y_2$]. So, $fd = d'f$.

So putting it all together, we get $f_*D = D'f_*$.

Answer 2

I am not quite convinced by Rotman's construction of the connecting homomorphism in the MV sequence. His construction doesn't make it explicit how a cycle in $H_n(X)$ is mapped to a cycle in $H_{n-1}(X_1 \cap X_2)$. (At least its not obvious to me how one can arrive at this explicit description using Rotman's contruction.)

Here is another approach to it (based on Hatcher).

Suppose you have a short exact sequence of chain complexes:

$$ 0 \to A_* \xrightarrow{i} B_* \xrightarrow{j} C_* \to 0 $$

then by zig-zag lemma, we get an LES in homology:

$$ \cdots \to H_n(A) \xrightarrow{i_*} H_n(B) \xrightarrow{j_*} H_n(C) \xrightarrow{\partial_*} H_{n-1}(A) \to \cdots $$

where $i_*$, $j_*$ are obvious. I will recall what the connecting homomorphism $\partial_* : H_{n}(C) \to H_{n-1}(A)$ does. You start with an element in $c\in C_n$, since $j$ is surjective, you can choose a cycle $b \in B_n$ such that $j(b) = c$. Now by the boundary map you move to $\partial b \in B_{n-1}$. Use commutativity and exactness to argue that this $\partial b$ lies in image of $i$. So, you can move to $a = i^{-1}(\partial b) \in A_{n-1}$.

More presicely, the map $\partial_*$ takes the homology class $[c]$ to the class $[i^{-1}\partial j^{-1}(c)]$. Of course, there are some things that need to be checked. Like, why is this map well-defined? We are making two choices: one is the choice of a representative for $[c]$ and the other is the choice of a pre-image $j^{-1}(c)$. It is an easy exercise to show that $[i^{-1}\partial j^{-1}(c)]$ is a unique homology class regardless of those choices. (Refer Bredon or Hatcher).

In the course of proving Excision or MV, you will prove the following result: the inclusion $\iota : C_n(X_1+X_2) \hookrightarrow C_n(X)$ is a chain-homotopy equivalence, where $C_n(X_1 + X_2)$ is the chain group of singular simplies which have their images in either entirely in $X_1$ or in $X_2$ (Munkres calls these small chains). Actually, the chain-homotopy inverse $\rho: C_n(X) \to C_n(X_1 + X_2)$ is given by barycentric subdivision, which takes a chain in $X$ and maps it to a small chain in $\{X_1, X_2\}$.

Now, consider the short exact sequence,

$$ 0 \to C_n(X_1 \cap X_2) \xrightarrow{i} C_n(X_1) \oplus C_n(X_2) \xrightarrow{j} C_n(X_1 + X_2) \to 0 $$

where the $i$ is signed inclusion $c \mapsto (c, -c)$ and the map $j$ just sums up two chains $(a, b) \mapsto a+b$.

Applying the zig-zag lemma, we obtain this LES:

$$ \cdots \to H_n(X_1 \cap X_2) \xrightarrow{} H_n(X_1) \oplus H_n(X_2) \xrightarrow{} H_n(X_1 + X_2) \xrightarrow{\partial_*} H_{n-1}(X_1 \cap X_2) \to \cdots $$

Since $\iota$ is a quasi-isomorphism, we get:

$$ \cdots \to H_n(X_1 \cap X_2) \xrightarrow{} H_n(X_1) \oplus H_n(X_2) \xrightarrow{} H_n(X) \xrightarrow{\partial_*} H_{n-1}(X_1 \cap X_2) \to \cdots $$

Now, using the construction of $\partial_*$ as in the zig-zag lemma above, we describe $\partial_*$ for this LES.

We start with a class $\alpha$ in $H_n(X)$ represented by a cycle $c$. By $\rho$ we can choose $c$ to be a small chain $c = a + b$, where $a$ is a chain in $X_1$ and $b$ in $X_2$. (Note that, while $c$ is a cycle $a$ and $b$ may not be cycles individually, but there sum is, so $\partial(a+b) = 0$ or $\partial a = -\partial b$).

Now, we apply $i^{-1}\partial j^{-1}$ as in zig-zag lemma: We can choose $j^{-1}(c) = (a, b)$ (we don't have worry about which pre-image to choose because in the proof of zig-zag lemma we show that the choice of pre-image has no change on the homology class). Then, we apply $\partial$ to get $(\partial a, \partial b)$. Finally, we invert the signed inclusion to get $\partial a$. Thus, $\partial_*\alpha = \partial a$.

Loosely, $\partial_*$ sends a cycle to boundary of one of its component after barycentric subdivision.

Now, suppose $\partial_*$ is the connecting homomorphism in the MV sequence for $Y$. We want to show that $f_*\partial_* = \partial'_* f_*$.

Let $\alpha \in H_n(X)$. Then $f_*\partial_*(\alpha) = f_*(\partial a)$.

Next, let $f_*(\alpha) \in H_n(Y)$. We can choose the small chain to be $f*(a) + f*(b)$ (note that this is not the only subdivision possible, but zig-zag lemma guarantees that that the choice of inverse under $j^{-1}$ will have no change on the resulting homology class). So, $\partial_*f_*(\alpha) = \partial f_*(a) = f_*(\partial a) = f_*\partial_*(\alpha)$.

So, $f_*\partial_* = \partial'_* f_*$.