Long exact sequence of a triple: working through the geometry Suppose $X$ is a topological space with subspaces $X \supset U \supset A$ such that $U$ deformation retracts onto $A$. We know that $H^*(X,U) \cong H^*(X,A)$--one way to see this is to take the long exact sequence of the triple $(X,U,A)$.
I was trying to get an intuition for why the map $(X,A) \to (X,U)$ induces an isomorphism on cohomology, so that I'm not blinded by the slickness of the  homological algebra, while missing what's going on. This is what I came up with: it's kind of convoluted, and I'd love to know if I'm overcomplicating the matter.
Edit: This stuff below has little to no usefulness; I put a reworked argument in my answer below.

The map $H^k(X,U) \to H^k(X,A)$ is injective as follows: if there were
  a relative $(X,U)$ cocycle $\varphi$ which was an $(X,A)$ coboundary
  (but perhaps not an $(X,U)$ coboundary), that would mean that
  $\varphi$ is the same as $C_{k}(X) \overset{\partial}{\to} C_{k-1}(X)
 \overset{\mu}{\to} R$ for a map $\mu$ that vanishes on $A$. (Here $R$
  is the ground ring). But since $\varphi$ vanishes on $U$, being an
  $(X,U)$ cocycle, this means that $\mu$ must vanish on chains in
  $C_{k-1}(U)$ which are boundaries. But then $\mu$ must vanish on all
  relative $(U,A)$ cycles, because they are all $(U,A)$ homologous to
  chains in $A$. If $\mu$ vanishes on all relative $(U,A)$ cycles, it
  descends to a map out of $\frac{C_{k-1}(U)}{\text{Stuff with boundary
 in $A$}} \cong \operatorname{Im} \partial_{k-1} \subset
 \frac{C_{k-2}(U)}{C_{k-2}(A)}$, and since
  $\frac{C_{k-2}(U)}{\operatorname{Im}\partial_{k-1}}$ is free and
  therefore $\operatorname{Ext}^1_R \left(
 \frac{C_{k-2}(U)}{\operatorname{Im}\partial_{k-1}}, C_{k-2}(U)
 \right)=0$, we can extend this map $\operatorname{Im}\partial_{k-1}
 \to R$ to a map $C_{k-2}(U) \to R$, and then we have expressed $\mu$
  as $C_{k-1}(X) \overset{\partial_{k-1}}{\to} C_{k-2}(X) \to R$. In
  other words, we have expressed $\varphi$ as a map that starts with two
  instances of boundary maps, and so $\varphi$ is zero. 
The map $H^k(X,U) \to H^k(X,A)$ is surjective as follows: if $\varphi
 \in C^k(X)$ vanishes on $A$ and on boundaries, then then its
  precomposition with $U \overset{i}{\hookrightarrow} X$ vanishes on
  boundaries, so it must vanish on all relative cycles, since any chain
  in $U$ differs from a chain in $A$ by a relative coboundary. Since
  $i^* \varphi$ vanishes on all relative cycles, it must factor as
  $C_{k}(U) \overset{\partial}{\to} C_{k-1}(U) \overset{\mu}{\to} R$ for
  a map $\mu$ that vanishes on $A$. Take any extension of $\mu$ to $X$
  and call it $\tilde{\mu}$. Now we have that $\varphi - \tilde{\mu}
 \circ \partial$ vanishes on all of $U$, and it still vanishes on
  boundaries, and it differs from $\varphi$ by a relative $(X,A)$
  coboundary.

 A: The geometry comes precisely from the long exact sequence of triple: $A \subset U \subset X$, so consider the map $C_n(X, A) \to C_n(X, U)$ by realizing a relative chains $\alpha$ in $X$, i.e., a chain with $\partial \alpha \subset A$, as a cycle relative to $U$, since $\partial\alpha \subset A \subset U$. This descends to a map $f : H_n(X, A) \to H_n(X, U)$.
Kernel of this map are represented by the relative cycles $\alpha$ in $(X, A)$ such that $\partial \alpha$ is null as a chain of $U$. If the kernel is nontrivial, we could find an $\alpha$ such that $\partial \alpha$ is null in $A$ yet non-null in $U$. This is impossible, as that'd give $A$ and $U$ different homologies in the same dimension: they are homotopy equivalent.
This is also obviously surjective, as given any chain $\beta$ in $X$ with $\partial \beta$ in $U$ (i.e., a relative cycle in $(X, U)$), I can homologoue $\beta$ by perturbing $\partial \beta \subset U$ into $r(\partial \beta) \subset A$ where $r : U \to A$ is the retraction at the end of our deformation retract. Since small homotopies leave homology classes alone, image of the homology class of this new cycle in $(X, A)$ by $f$ is precisely $[\beta]$.
$f$ is an isomorphism, as desired. The proof is analogous for cohomology, but geometrically less apparent. If you are lazy, just use universal coefficient theorem to conclude (if a chain-level map induces isomorphism on homology, then it's dual must also induce isomorphism on cohomology)
A: The argument I was trying to find was the following:
Consider the map $H^k(X,U) \to H^k(X,A)$. To show it is surjective, suppose that $\psi \in C^k(X)$ is a cocycle vanishing on $A$. Any chain $\sigma \in C_k(U)$ is related to a chain $\tilde{\sigma} \in C_k(A)$ by $\partial (P\sigma) = \sigma - \tilde{\sigma} - P(\partial \sigma)$, where $P$ is the prism operator for the deformation retraction of $U$ onto $A$. Since $\psi$ vanishes on boundaries and on $A$, $\psi(\sigma) = \psi(P(\partial \sigma))$ for $\sigma \in U$, and so $\psi$ acts on $U$ as a coboundary. Therefore $\psi - \psi \circ P \circ \partial$ is zero on $U$, and $\psi$ is cohomologous to a function vanishing on $U$.
To show that $H^k(X,U) \to H^k(X,A)$ is injective, suppose that $\varphi \in C^k(X)$ is a cocycle vanishing on $U$. We show that if it is an $(X,A)$ coboundary, then it is an $(X,U)$ coboundary. If it is an $(X,A)$ coboundary, then there is a cochain $\mu \in C^{k-1}(X)$ vanishing on $A$ such that $\varphi = \mu \circ \partial$. Now $\mu$ vanishes on $\partial(C_k(U))$ by construction, therefore its restriction to $U$ is a cocycle in $C_{k-1}(U)$ which vanishes on $A$, so $\left. \mu \right|_U = \left.\mu \right|_U \circ P \circ \partial$, as in the last paragraph. Extend $P$ to $C_{k-1}(X)$ however you like, and now $\varphi = \mu \circ \partial = (\mu - \mu \circ P \circ \partial) \circ \partial$, and $\mu - \mu \circ P \circ \partial$ vanishes on $U$. 
