Is this Mayer-Vietoris sequence motivic? Suppose $Y$ is a variety defined over $\mathbb{Q}$ and $pt$ is a rational point of $Y$. Let $\pi:X \rightarrow Y$ be the blow up of $Y$ at $pt$ and $D$ be the exceptional divisor. For simplicity let's assume both $X$ and $D$ are smooth. Let $CD$ be the cone of $D$ defined to be
\begin{equation}
D \times I/D \times \{0\}
\end{equation}
where $I$ is the unit interval $[0,1]$. Now let $Y'$ be
\begin{equation}
Y'= X \sqcup CD /D \times \{ 1\}
\end{equation}
then the cw complex $Y'$ is homotopic to $Y$, which the morphism $\pi$ is homotopic to the inclusion $i:X \rightarrow Y'$. Then $(CD,X)$ is a cover of $Y'$ while the intersection of $CD$ and $X$ is just $D \times \{ 1\} \simeq D$. The cover $(CD,X)$ of $Y'$ induces a Mayer-Vietoris sequence
\begin{equation}
\cdots \rightarrow H^n(Y',\mathbb{Q}) \xrightarrow{\pi^*} H^n(X,\mathbb{Q}) \xrightarrow{j^*}H^n(D,\mathbb{Q}) \xrightarrow{\delta}H^{n+1}(Y',\mathbb{Q}) \rightarrow \cdots
\end{equation}
where $j$ is the inclusion morphism $D \rightarrow X$. 
The morphism $\pi^*$ and $i^*$ is acutually motivic since they are induced by morphisms in the category of $\mathbb{Q}$-varieties. Is the morphism $\delta$ in this long exact sequence motivic?
 A: I'm not sure if that's exactly what you mean by "being motivic", but this long exact sequence comes from a triangle in Voevodsky's category of (integral) motives $DM(\mathbb Q)$. 
More generally if $X$ is a $\mathbb Q$-scheme of finite type, $Z\subset X$ a closed subscheme, $\pi : Y \to X$ the blowup at $Z$, and $E=\pi^{-1}(Z)$, then there is a triangle
$$
M(E) \to M(Y)\oplus M(Z) \to M(X) \to M(E)[1]
$$
in $DM(\mathbb Q)$. This is proved in Mazza–Voevodsky–Weibel's Lecture notes on motivic cohomology, see equation (14.5.3). It can also be viewed a formal consequence of the six-functor formalism for $DM(-)$, as described by Cisinski and Déglise in their paper Integral mixed motives in equal characteristic.
Added later: Let me sketch the abstract proof, which works in the $\ell$-adic context as well. The proof only uses that $\pi:Y\to X$ is a proper map that restricts to an isomorphism over the complement of $Z$ (an "abstract blow-up"). The triangle comes from a homotopy cocartesian square $1_X=M(Y) \coprod_{M(E)}M(Z)$ in $DM(X)$. Here for $f: X' \to X$ of finite type, we define $M(X')=f_!f^!(1_X)$. Let $i: Z\to X$ be the inclusion and $j: U\to X$ the open complement. Then the pair of functors $(i^!,j^*)$ is conservative. But applying either functor to the given square, using the proper base change theorem, gives a square which is cocartesian for formal reasons (opposite sides are equivalences).
