On the hilbert polynomial of a coherent sheaf over a projective scheme This is a lemma from the book by Huybrechts and Lehn.
Let, $$
be a projective scheme over a field $\mathbb K$. Let, $\mathcal O(1)$ be an ample line bundle on $$, then the Hilbert polynomial $()$ is given by $↦( \otimes \mathcal O())$
In this set up the lemma says that 

: Let $$ be a coherent sheaf of dimension $$ and let $H_1,...,H_d \in|\mathcal O(1)|$ be an $$-regular sequence. Then $P(E, m) = \chi(E ⊗ \mathcal{O}(m)) =\Sigma_{i=0}^{d}\chi(E|_{\cap_{j\leq i }H_j}){m+i-1\choose i}$.
we have the following proof of it 
I
have the following doubts in understanding the proof of this :
$(1)$ It's said that for $d=0$ the assertion is trivial.What is the right hand side when $d=0$?(In this case I realize that $E$ is supported on points, but how does that imply the desired result?)
$(2)$ from $0 \to \mathcal I_H  \to \mathcal O_X \to i_*(\mathcal O_X) \to 0$ ...$(a)$, how can one have $ 0 \to E(m-1)  \to  E(m) \to E(m)|_H \to 0$ ? I know the map $E \to E \otimes \mathcal O(1)$...$(b)$ is injective ,but it's not quite clear why we end up having the desired S.E.S?(More precisely how the sequence derived from $(a)$ remains left exact using the injective map $(b)$?)
$(3)$ Why $E(m)|_H$ has dimension $d-1$?
$(4)$ If we denote $f(m) = \chi(E ⊗ \mathcal{O}(m)) - \Sigma_{i=0}^{d}\chi(E|_{\cap_{j\leq i }H_j}){m+i-1\choose i}$, then we have $f(m)-f(m-1)=0$. But,For $m=0$, I don't see how $\Sigma_{i=0}^{d}\chi(E|_{\cap_{j\leq i }H_j}){m+i-1\choose i} = \chi(E)$?
Any help from anyone will be appreciated 
 A: *

*If $d = 0$, then the right hand side is just
$$\chi(E) \cdot \binom{m-1}{0} = \chi(E).$$
This is consistent with the fact that if $E$ is supported on finitely many points, then $E = \bigoplus_{P \in X} E_P$, i.e. it is the finite direct sum of skyscraper sheaves of its stalks. Twisting does not change the stalks, so $E(m) = E$ for all $m \in \mathbb{Z}$.

*This is true because if $D$ is any effective Cartier divisor, then $I_D = \mathcal{O}(-D)$. In this case, $H$ is a hyperplane section, so $I_H = \mathcal{O}(-1)$. Tensoring the defining short exact sequence for $H$ $$ 0 \to \mathcal{O}_X(-1) \to \mathcal{O}_X \to \mathcal{O}_H \to 0 $$
with $E$ leads to the sequence
$$0 \to E(-1) \to E \to E|_H \to 0.$$
This sequence is exact on the left, because $H$ is regular with respect to $E$.

*The regularity implies that $H$ contains none of the associated points of $E$, see the paragraph directly under Definition 1.1.11. But this means that $H$ misses all generic points of the irreducible components of the support of $E$, see stacks-project, associated primes. So $H$ does not contain any component of $\operatorname{Supp}(E)$, hence the dimension of $E$ drops by $1$.

*We have $$f(m) = \chi(E(m)) - \sum_{i = 0}^{d} \chi(E|_{\bigcap_{j \leq i}H_j})\binom{m + i - 1}{i}.$$
For $m = 0$ the binomial coefficients become $\binom{i-1}{i}$ which equals zero for $i \geq 1$, and equals $1$ for $i = 0$ (at least I hope this is the intention, but I dont see any other way how the statement could be correct).
So $f(0) = \chi(E) - \chi(E) = 0$.
