$\delta:H^n(\{0\}\times Y; R)\to H^{n+1}(I\times Y, \partial I, R)$ is an isomorphism. 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 $\delta:H^n(\partial I\times Y;R)\to H^{n+1}(I\times Y, \partial I\times Y; R)$.

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 $[\phi]\in H^n(\partial I\times Y; R)$, 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 $\partial I\times Y$. 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 $\phi(\gamma)$ 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.
 A: Given the long exact sequence
    $$0\rightarrow H^n(I\times Y;R)\rightarrow H^n(\partial I\times Y;R)\xrightarrow{\delta}H^{n+1}(I\times Y,\partial I\times Y;R)\rightarrow0.$$
Notice that $$H^n(\partial I\times Y;R)\cong H^n(\{0\}\times Y)\oplus H^n(\{1\}\times Y).$$
    Use $M,M_0,M_1$ to respectively denote $H^n(Y;R),H^n(\{0\}\times Y;R),H^n(\{1\}\times Y;R)$ and use $i_0,i_1$ to respectively denote the inclusion from $\{0\}\times Y,\{1\}\times Y$ to $I\times Y$. Then we have
    $$M_0\xrightarrow[\cong]{i_0^\ast}M\xleftarrow[\cong]{i_1^\ast}M_1.$$
    They are isomorphisms because $I\times Y\sim\{0\}\times Y\sim\{1\}\times Y$. And the dual map $i_0^\ast,i_1^\ast,i^\ast$ are all restrictions of $R$-module homomorphisms. Thus $i^\ast(\alpha)=(i^\ast_0(\alpha),i^\ast_1(\alpha))\in M_0\oplus M_1$. $i_0^\ast,i^\ast_1$ being isomorphisms implies that the submodule $\mathrm{im} i^\ast\subseteq M_0\oplus M_1\cong M\oplus M$ has the form (if $\mathrm{im} i^\ast$ regarded as the submodule of $M\oplus M$)
    $$\mathrm{im} i^\ast=\{(\alpha,\alpha)|\alpha\in M\}.$$
    The subtle conclusion from this is that
    \begin{equation}\label{eg3.11}
 M_0\oplus M_1=M_0\oplus\mathrm{im} i^\ast.\tag{$\ast$}
 \end{equation}
    Now let's go back to the short exact sequence. By exactness $\ker\delta=\mathrm{im} i^\ast$, and $\delta$ is surjective. Then with the help of (\ref{eg3.11}) it is easy to prove that $\delta|_{M_0}$ is an isomorphism.
(I think (\ref{eg3.11}) is the key point.)
