TITLE: question about a line in Hatcher about long exact sequence in cohomology
QUESTION [6 upvotes]: On p.210 Example 3.11 of Hatcher's Algebraic Topology book, he makes the assertion in the following screenshot - where $Y$ is any space, and $I$ is the unit interval.

My question is why this is the case...for example why is the map $H^n(I \times Y; R) \to H^n(\partial I \times Y; R)$ in the long exact sequence injective?
Here is my attempt:
If we have a cocycle $c\in C^n(I \times Y)$ representing a class $[c] \in H^n(I \times Y; R)$ that maps to $0$ in $H^n(\partial I \times Y; R)$, that means the image $i(c)$ of $c$ in $C^n(\partial I \times Y; R)$ is a coboundary, i.e there is a $d \in C^{n-1}(\partial I \times Y; R)$ such that $\delta d=i(c)$. We can extend/lift $d$ to $D \in C^{n-1}(I \times Y; R)$ by defining $D$ to vanish on $n-1$-simplices in $I \times Y$ that are not contained in $\partial I \times Y$. Given a $n$-simplex in $I \times Y$, it need not be the case that $\delta D =c$, since for a $n$-simplex in $I \times Y$ that is not contained in $\partial I \times Y$, the left hand side $\delta D$ is zero, while the right hand side $c$ need not be zero, so this argument doesn't seem to work.

REPLY [5 votes]: Consider the map $p_0:I\times Y\to\partial I\times Y, (t,y)\mapsto (0,y)$, then if $i:\partial I\times Y\hookrightarrow I\times Y$ is the inclusion, then $i\circ p_0$ is the map $(t,y)\mapsto (0,y)$ and is clearly homotopic to $\mathrm{id}_{I\times Y}$, hence $p_0^*i^*=\mathrm{id}$ in cohomology. This shows that $i^*$ is injective, and by exactness, if $j$ is the inclusion of pairs $j:(I\times Y,\emptyset)\hookrightarrow(I\times Y,\partial I\times Y)$, then $\mathrm{Ker}~i^*=\mathrm{Im}~j^*=0$, so $j^*=0$ and the connecting homomorphism $\delta$ is onto because $\mathrm{Im}~\delta=\mathrm{Ker}~j^*$. You get short exact sequences
$$0\to H^{n}(I\times Y)\xrightarrow{\;i^*\;} H^{n}(\partial I\times Y)\xrightarrow{\;\delta\;}H^{n+1}(I\times Y,\partial I\times Y)\to 0$$
and the first arrow is split (through $p_0^*$).