algebraic-topologyexact-sequencehomology-cohomology
I am trying to understand a long exact sequence from some notes. $I=[0,1] $ and $X$ any topological space, We look at the long exact sequence for the pair $(I\times X,\partial I\times X)$
$\dotsb\xrightarrow{0}H^k(I\times X)\rightarrow H^k(\partial I\times X)\xrightarrow{\delta}H^{k+1}(I\times X,\partial I\times X)\xrightarrow{0}H^{k+1}(I\times X)\rightarrow\dotsb$
I can't understand why the mentioned $0$-maps are $0$. An explanation would be very helpful.
I think you can, but the application of your spectral sequence is rather forced. This is what I mean.
First consider the "braid" diagram
$\hskip0.3in$
where the blue, purple, and black sequences are the long exact sequence for their respective pairs, $\beta = \kappa \circ \theta$, and $\pi,\eta,\lambda$ are induced by inclusions of pairs. We want to show the orange sequence is also exact. Note that it is already a complex by the commutativity of the diagram, and because $\eta \circ \pi$ also factors through $H_{q+2}(Y,Y)=0$.
Remark. Using a spectral sequence actually makes this argument really confusing. It's easier to directly deduce the exactness of the orange sequence using the braid lemma, which is §26, Exc. 1 in Elements of algebraic topology by Munkres, which is just a simple diagram chase. Of course, there might be a spectral sequence proof of this lemma.
Proof of exactness. We use the spectral sequence associated to the filtration of our space $X \supset Y \supset Z$, as stated in Homotopic topology by Fomenko/Fuchs/Gutenmacher, §18. Let $X_2 = X$, $X_1 = Y$, $X_0 = Z$, and $X_{-1} = \emptyset$. The spectral sequence then states that if on the $E^1$ page we have \begin{equation*}\DeclareMathOperator{\im}{im} E^1_{p,q} = \begin{cases} H_q(Z) & \text{if}\ p=0\\ H_{q+1}(Y,Z) & \text{if}\ p=1\\ H_{q+2}(X,Y) & \text{if}\ p=2\\ 0 & \text{otherwise} \end{cases} \end{equation*} this converges on the $E^\infty$ page to \begin{equation*} E^\infty_{p,q} = \frac{\im(H_{p+q}(X_p) \to H_{p+q}(X))}{\im(H_{p+q}(X_{p-1}) \to H_{p+q}(X))}. \end{equation*}
To calculate the spectral sequence, we note that on the $E^1$ page, the differentials in the $q$th rows are of the form \begin{equation*} 0 \longleftarrow H_q(Z) \overset{\gamma}{\longleftarrow} H_{q+1}(Y,Z) \overset{\beta}{\longleftarrow} H_{q+2}(X,Y) \longleftarrow 0. \end{equation*} Thus, on the $E^2$ page we have \begin{equation*} E^2_{p,q} = \begin{cases} \dfrac{H_q(Z)}{\im\gamma} & \text{if}\ p=0\\ \dfrac{\ker\gamma}{\im\beta} & \text{if}\ p=1\\ \ker\beta & \text{if}\ p=2\\ 0 & \text{otherwise} \end{cases} \end{equation*} on which the (non-trivial) differentials $E_{2,q}^2 \to E_{0,q+1}^2$ are of the form \begin{equation*} 0 \longleftarrow \frac{H_{q+1}(Z)}{\im\psi} \longleftarrow \ker\beta \longleftarrow 0, \end{equation*} and this map is the map which takes an element in $\ker\beta$, maps it down through $\theta$, and lifts it up to $H_{q+1}(Z)$. Now since all differentials are trivial on the $E^3$ page, we have that \begin{equation*} E^\infty_{0,q+1} = E^3_{0,q+1} = \frac{H_{q+1}(Z)/\im\psi}{\im(\ker\beta)} \overset{\sim}{\longrightarrow} \im\omega \end{equation*} where this isomorphism is induced by $\omega$; \begin{equation*} E^\infty_{1,q} = E^3_{1,q} = \frac{\ker\gamma}{\im\beta} \overset{\sim}{\longrightarrow} \frac{\im \tau}{\im \omega} \end{equation*} where this map takes an element in $\ker\gamma$, pulls it back to $H_{q+1}(Y)$, then maps it to $H_{q+1}(X)$; \begin{equation*} E^\infty_{2,q} = E^3_{2,q} = \ker\left( \ker\beta \to \frac{H_{q+1}(Z)}{\im\psi} \right) \overset{\sim}{\longleftarrow} \frac{H_{q+2}(X)}{\im \delta} \end{equation*} where this map is induced by $\alpha$. Then, exactness at $H_{q+1}(X,Z)$ follows by a diagram chase using the isomorphism for $E^\infty_{0,q+1}$, at $H_{q+2}(X,Y)$ by the isomorphism for $E^\infty_{2,q}$, and at $H_{q+1}(Y,Z)$ by the isomorphism for $E^\infty_{1,q}$. $\blacksquare$
Remark. Note that in the spectral sequence argument, the spectral sequence doesn't help you too much since you end up having to understand all the maps in the spectral sequence anyway, and do a bit of diagram chasing at the end, although the diagram chase is simpler than that you would have to do for the proof of the braid lemma.
EDIT. I tried to turn this into the language of exact couples; here are the diagrams involved. First fix some notation. Let $Z \overset{i}{\hookrightarrow} Y \overset{j}{\hookrightarrow} Z$, and let $$H(Z) \overset{i_*}{\longrightarrow} H(Y) \overset{q_{Y,Z}}{\longrightarrow} H(Y,Z) \overset{\delta_{Y,Z}}{\longrightarrow} H(Z)[1]\\ H(Y) \overset{j_*}{\longrightarrow} H(X) \overset{q_{X,Y}}{\longrightarrow} H(X,Y) \overset{\delta_{X,Y}}{\longrightarrow} H(Y)[1]\\ H(Z) \overset{(j \circ i)_*}{\longrightarrow} H(X) \overset{q_{X,Z}}{\longrightarrow} H(X,Z) \overset{\delta_{X,Z}}{\longrightarrow} H(Z)[1]\\ H(Y,Z) \longrightarrow H(X,Z) \longrightarrow H(X,Y) \overset{\beta}{\longrightarrow} H(Y,Z)[1]$$ where $\beta = q_{Y,Z} \circ \delta_{X,Y}$. Now consider the exact couple
$\hskip0.75in$
The first derived couple is
$\hskip0.75in$
and the second derived couple is
$\hskip0.75in$
which does indeed give the isomorphisms claimed above, but this doesn't simplify the argument much…
I will try to answer the question in a way which you probably did not expect: Rather than giving an alternative quicker solutition I am trying to convince you that your solution is rather clear.
First of all, I would define $A=X\times ([0,1/3)\cup (1/3,1])$ and $A=X\times ([0,2/3)\cup (2/3,1])$ and then choose the orientation of the homology to be induced by the inclusion of $X\times \{1/2\}$. For me, this makes it even more obvious why the map $H_*(X)\oplus H_*(X)$ should be given by $F=\begin{pmatrix} 1&1\\1 & f_*\end{pmatrix}$. (Note: The sign in the last column differs from your solution, but this is up to personal preference of the orientation of the homology groups.)
By changing the basis and writing $(x_1,x_2)\in H_*(X)\oplus H_*(X)$ as $(x_1+x_2,x_2)$ and $(y_1,y_2)\in H_*(X)\oplus H_*(X)$ as $(y_1,y_1-y_2)$ one then gets $F$ written with respect to the new basis as $F'=\begin{pmatrix} 1&0\\0&1-f_*\end{pmatrix}$. Because the sequence is exact, we then can omit the first copy in each of the two terms of the l.e.s. and obtain the desired result.
In other words, I believe your solution is already the correct one. However, sometimes it is even harder to understand why a solution is quite simple than it is to write it down in the first place.
Best Answer
The map $H^k(I\times X)\rightarrow H^k(\partial I\times X)$ is induced by the inclusion $i:\partial I\times X\hookrightarrow I\times X$. You can construct a map $s$ in the opposite direction as the composite
$$s:I\times X\xrightarrow{pr}I\times X\xrightarrow{in_0}\partial I\times X.$$
Check that the composite $i\circ s:X\times I\rightarrow X\times I$ is homotopic to the identity. It follows that on cohomology you have $(i\circ s)^*=(id_{I\times X})^*=id$.
Now we see that the map $i^*$, which is the first map in your exact sequence, has a left inverse $s^*$, since
$$s^*i^*=(i\circ s)^*=id$$
where we have used functorality.
It follows that $i^*:H^k(I\times X)\rightarrow H^k(\partial I\times X)$ is injective for all $k\geq 0$, and so by exactness of the sequence that the map preceeding it, $H^{k-1}(I\times X,\partial I\times X)\rightarrow H^k(I\times X)$, is zero. You get this from the definitions using $ker(i^*)=0$