Long exact sequence of $(I \times Y, \partial I \times Y)$. There's a section in chapter 3 in Hatcher's Algebraic Topology where he talks about the long exact sequence of the pair $(Y \times I, Y \times \partial I)$, where $Y$ is any topological space. The claim is that the map 
$$H^k(I \times Y, \partial I \times Y) \to H^k(I \times Y)$$
in the long exact sequence is zero for all $k$. I wanted to see this geometrically, instead of losing the geometry in the homological algebra. I think I am satisfied, but I just wanted to check that my explanation wasn't a) wrong, or b) overcomplicating something simpler.
Here's my explanation: 

Suppose $\varphi$ is a cochain in $C^k(I \times Y)$ which vanishes on
  $\partial I \times Y$. If it is a cocycle (i.e. if it vanishes on
  boundaries), then it must in fact vanish on all cycles, because any
  cycle in $I \times Y$ is homologous to a cycle in $\{0\} \times Y
 \subset \partial I \times Y$--namely, its image under the deformation retract of $I \times Y$ onto $ \{0\} \times Y$.  Now if $\varphi$ is zero on all cycles, it descends to a map $\frac{C_k(I \times Y)}{\ker \partial_k} \to R$ or $\operatorname{Im}\partial_k \to R$, where $R$ is the ground ring. Now since $\frac{C_{k-1}}{\operatorname{Im}\partial_k}$ is free, $\operatorname{Ext}^1_R(\frac{C_{k-1}}{\operatorname{Im}\partial_k}, \operatorname{Im}\partial_k)=0$, so any map $\operatorname{Im}\partial_k \to R$ extends to a map $C_{k-1}(I \times Y) \to R$, and therefore $\varphi$ is a coboundary. 
Note that $\varphi$ need not be a relative coboundary in $C^k(I \times
 Y, \partial I \times Y)$, because the aformentioned map $C_{k-1}(I
 \times Y) \to R$ need not be zero on $C_{k-1}(\partial I \times Y)$.

Edit: I made a mistake; $\frac{C_{k-1}}{\operatorname{Im}\partial_k}$ isn't necessarily free, so I need to fix this.
Edit 2: All this above stuff has little to no use; I posted a simpler explanation in an answer below.
 A: I am writing out my comment as an answer to get this off the unanswered queue.

Here's a (in my opinion) simple explanation for the claim: 
$H^n(I \times Y, \partial I \times Y) \to H^n(I \times Y)$ is induced from the inclusion $i : (I \times Y, \emptyset) \hookrightarrow (I \times Y, \partial I \times Y)$. Consider the inclusion $j : (\{0\} \times Y, \emptyset) \hookrightarrow (I \times Y, \emptyset)$. $i \circ j$ takes everything inside $\{0\} \times Y \subset \partial I \times Y$, which is precisely what we are rel-ing out, so $H^n(j) \circ H^n(i) = H^n(i \circ j) = 0$. But $j$ is an isomorphism on cohomology, as there is a deformation retraction $j : I \times Y \to \{0\} \times Y$. Hence, $H^n(i) = 0$. 
Geometrically, $I \times Y$ deformation retracts to $\{0\} \times Y$ but under this retraction $\partial I \times Y$ maps down to all of $\{0\} \times Y$. So cohomologically image of $i_\#$ (at the cochain level) is indistinguishable from the image of $C^n(\{0\} \times Y, \{0\} \times Y) \to C^n(\{0\} \times Y)$ induced from the inclusion $(\{0\} \times Y, \emptyset) \hookrightarrow (\{0\} \times Y, \{0\} \times Y)$, elements of which are of course all nullcohomologous. 
A: I appreciate Balarka's helpful answer, especially since it wasn't super clear what I was looking for. What I wrote below is basically the same answer, except made a little more explicit and (perhaps) geometric.
To say that the given map $H^k(I \times Y, \partial I \times Y) \to H^k(I \times Y)$ is zero for all $k$ is to say that any cochain $\varphi \in C^k(I \times Y)$ which vanishes on $\partial I \times Y$ and vanishes on boundaries (i.e. a cocycle) must then factor through the boundary map. Now $I \times Y \overset{r}{\to} \{0\} \times Y$ is a deformation retract with homotopy $i \circ r \simeq \text{Id}_{I \times Y}$. Let $P$ be the prism operator corresponding to this homotopy.
Any chain $\sigma \in C_k(I \times Y)$,  is related to $\tilde{\sigma} = (i \circ r)_* \sigma$ by 
$$\partial (P\sigma) = \sigma - \tilde{\sigma} - P(\partial \sigma).$$
Therefore, for our $\varphi$ which vanishes on boundaries and on $\partial I \times Y$, we have
$$0 =\varphi(\partial (P\sigma)) = \varphi(\sigma
 - \tilde{\sigma} - P(\partial \sigma)) \implies \varphi(\sigma) = \varphi(P(\partial \sigma)).$$ 
Therefore $\varphi$ factors through the boundary map: $\varphi = \varphi \circ P \circ \partial$. 
