In the last paragraph of pg. 210 of Hatcher's Algebraic Topology, the following is mentioned:


Let $Y$ be a topological space and $I$ be the closed interval $[0, 1]$. Consider the long exact sequence of cohomology (with coeff. in a ring $R$) for the pair $(I\times Y, \partial I\times Y)$. In this long exact sequence we have a map $\delta:H^n(\partial I\times Y;R)\to H^{n+1}(I\times Y. \partial I\times Y; R)$.


Question. It is said in the text that the map $\delta$ is an isomorphism when restricted to the copy of $H^n(Y, R)$ in $H^n(\partial I\times Y; R)$.


I am unable to give a proof of this. At least not an economical one.


Here is what I thought:
First let us see what the map $\delta$ is.
For any $[\phi]\in H^n(\partial I; R)$, where $\phi$ is an $n$-cocycle, we define $\delta[\phi]$ as follows: Let $\Phi:C_n(I\times Y; R)\to R$ be defined by extending $\phi$ in the obvious way, we just have $\Phi$ take the value $0$ on simplices not in $I\times Y$. Then $\delta[\phi]=[\delta\Phi]$. This makes sense because $\delta\Phi$ does vanish on $n+1$-chains in $\partial I\times Y$.


Now for my "proof". For the sake of understanding let us take $n=1$ and suppose there is a cocycle $\phi\in C_1(\{0\}\times Y; R)$ such that $\delta[\phi]=0$.
We want to show that $[\phi]=0$, that is $\phi(\sigma)$ depends only on the end points of the path $\gamma$ for all $\gamma$.
Since $[\delta\Phi]=0$, there is $\Psi\in C_1(I\times Y, \partial\times Y;R)$ such that $\delta\Psi=\delta\Phi$.





In the figure above, $\gamma$ is a singular $1$-simplex in $Y$, which is one of the bundaries of the singular $2$-simplex $\sigma$ in $I\times Y$.
Since $\delta\Phi=\delta\Psi$, we have $\delta\Phi(\sigma+\tau)=\delta\Psi(\sigma+\tau)$.
Using the fact that $\Psi$ vanishes on $1$-simplices in in $\partial I\times Y$ and that $\Phi$ vanishes on $1$-simplices in $Y$ which lies outside $\{0\}\times Y$, we get
$$\Phi(\gamma) = \Psi(\text{such of the two "vertical" $1$-simplices in the diagram})$$


Thus $\Phi$ depends only on the end points of $\gamma$.
Using $\Phi|_{\{0\}\times Y}=\phi$, we have $[\phi]=0$ and we are done.


Of course, this is nor a formal proof, and this is done only for $n=1$.
But the same idea can be used to give a proof for any $n$.


Can somebody please give a better proof.