To Prove. Let $Y$ be a topological space and $I$ be the closed unit interval. Then the long exact sequence (for cohomology) of the pair $(I\times Y, \partial I\times Y)$ breaks up into split short exact sequences$$0\rightarrow H^n(I\times Y; R)\rightarrow H^n(\partial I\times Y; R)\xrightarrow{\delta} H^{n+1}(I\times Y, \partial I\times Y; R)\rightarrow 0$$

I am able to show that the map $H^n(I\times Y; R)\rightarrow H^n(\partial I\times Y; R)$ is injective. This is because if $j_0:\{0\}\times Y\to \partial I\times Y$ and $i_0: \{0\}\times Y\to I\times Y$ denote the inclusion maps, then the composite $H^n(I\times Y; R)\rightarrow H^n(\partial I\times Y; R)\xrightarrow{j_0^*} H^n(\{0\}\times Y; R)$ is $i_0^*$, which is an isomorphism since $i_0$ is a homotopy equivalence. So the long exact sequence breaks up into short exact sequences as above.

But I fail to see why the sequence splits. Can someone please help? Thank you.