To understand a long exact sequence related to Cohomology

Question

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.

Answer 1

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 $ker(i^*)=0$

Answer 2

Let $j : \partial I \times X \to I \times X$ denote inclusion. We show that $j^* : H^k( I \times X) \to H^k(\partial I \times X)$ has trivial kernel. By exactness this means that $H^{k}(I\times X,\partial I\times X)\xrightarrow{}H^{k}(I\times X)$ has trivial image, i.e. is the zero map.

Let $p : I \times X \to X$ denote projection; it is a homotopy equivalence. Moreover let $i_0 : X \to \partial I \times X, i_0(x) = (x,0)$, and $r : \partial I \times X \to X$ be the projection. We have $r i_0 = id$, thus $i_0^* r^* =id$. Therefore $r^*$ is a monomorphism. We also have $p j = r$, thus $j^* p^* = r^*$. Since $p^*$ is an isomorphism, we see that $j^*$ is a monomorphism, thus has trivial kernel.