Linearly Equivalence of pullback of divisors Let $X_0=\mathbb{P}^2$ and $\eta: X_r \mapsto X_0$ be the blow-up of $p_1,\cdots, p_r$, where $p_1 \in X_0$. In a paper I am reading, the author states the following: If $C\subset X_r$ is an irreducible curve that is equal to the strict transform of some curve $D\subset \mathbb{P}^2$, then $C$ must be linearly equivalent to $$dL-\sum_i m_iE_i$$ where $d$ is the degree of $D$,  and $L$ and the $E_i$'s are given in the Picard group decomposition: $\text{Pic}(X_r)=\mathbb{Z}L \oplus \mathbb{Z}E_1\oplus\cdots \oplus\mathbb{Z}E_r$. 
The author stated that this is due to the fact that one has $\pi^\ast(D)=\tilde{D} + m_p E$ where $\tilde{D}$ is the strict-transform of $D$. However, I am unable to see why this formula leads to the linear equivalence stated above. Can any one help enlighten me? THank you very much.
 A: To simplify, I'll assume $r=1$ and the field $k$ has characteristic zero (maybe unnecessary). 
Seen as Cartier divisor $D$ is given by $(U_i,\{f_i=0\})$ for an affine open covering $\mathbb{P}^2 = \cup U_i$. Since $\pi\colon X_1\backslash E \to \mathbb{P}^2\backslash \{p_1\}$ is an isomorphism, we restrict to a neighborhood of $p_1$. 
Let $x,y$ be generators of the maximal ideal of the local ring $\mathcal{O}_{\mathbb{P}^2,p_1}$ and let $f(x,y)=0$ be a local equation for $D$. Then 
$$
f(x,y) = \sum_{i+j \geq m} a_{ij}x^iy^j
$$
where $m=m_{p_1}(D)$ is the multiplicity. On the other hand, the blow-up is locally given by two charts $V_1,(x,t)$ and $V_2,(u,y)$ and 
$$
\pi(x,t) = (x,xt), \,\pi(u,y) = (uy,y).
$$
Note that $E\cap  V_1 = \{ x=0\}$ and $E\cap V_2 = \{y=0\}$. It follows that the pull back of $D$ is defined on $V_1$ by
$$
f\circ \pi (x,t) = \sum_{i+j \geq m} a_{ij}x^{i+j}t^{j} = x^mg(x,t)
$$
where $x$ does not divide $g$ and $\{g=0\}$ is the equation of the strict transform of $D$. On the other chart we have the same.
Hence $\pi^\ast(D) = \bar D+ mE$.
