Global sections of a tangent sheaf of a blown-up surface. Let $V$ be a smooth projective algebraic surface and let $\pi\colon V'\to V$ be a blowup of a point $p\in V$. I would like to ask if the following is true:
$h^{0}(V,\Theta_{V})=h^{0}(V', \Theta_{V'})$,
where $\Theta_{X}$ denotes the tangent sheaf to $X$. I am particularly interested in the case of rational surfaces. 
As far as I know, we have $\pi_{*}\Theta_{V'}=\mathfrak{m}_{p}\cdot \Theta_{V}$ and $\pi^{*}\Theta_{V}=\Theta_{V'}\langle E \rangle (E)$, where $E$ is the exceptional curve of $\pi$ and $\Theta_{X}\langle D \rangle$ denotes the subsheaf of $\Theta_{X}$ consisting of vector fields tangent to a divisor $D$. However, I have no intuition on what the functors $\pi^{*}$ and $\pi_{*}$ do with the sheaf cohomology (and how to apply it to my original question). If someone knows a comprehensive guide I can learn such things, I would be very grateful for a reference.
Thank you for any help.
 A: This is not true. For an example with rational surfaces, one can check that if $X_0=\mathbb P^2$ and $X_i$ is the blowup of $X_0$ at $i$ points in general position, then
$H^0(T_{X_0})=8$,
$H^0(T_{X_1})=6$,
$H^0(T_{X_2})=4$,
$H^0(T_{X_3})=2$, and $H^0(T_{X_i})=0$ for $i>3$.
One way to see why this is not true is as follows: Let $f: X\to Y$ be the blowup of $Y$ at a point. The cotangent sequence for the blowup is 
$$
0\to f^* \Omega_Y \to \Omega_X \to \Omega_{X/Y}\to 0,
$$
with $\Omega_{X/Y}$ supported on the exceptional divisor $E$.
Dualizing and using that $\mathcal H\mathrm{om}(\Omega_{X/Y},\mathcal O_X)=0$ we get
$$
0\to T_X\to f^* T_Y\to \mathcal E\mathrm{xt}(\Omega_{X/Y},\mathcal O_X)\to 0.
$$
One can then check
$\mathcal E\mathrm{xt}(\Omega_{X/Y},\mathcal O_X) = \mathcal O_{E}(-E)=\mathcal O_{\mathbb P^1}(1)$.
Applying $H^0(-)$ and using that $H^0(f^* T_Y)=H^0(T_Y\otimes f_* \mathcal O_X)=H^0(T_Y)$ (since $f_* \mathcal O_X=\mathcal O_Y$), we get
$$
0\to H^0(T_X)\to  H^0(T_Y)\to H^0(\mathbb P^1)\to H^1(T_X)\to\cdots.
$$
Thus $H^0(T_X)=H^0(T_Y)$ if and only if $H^0(T_Y)\to H^0(\mathbb P^1)$ is the zero map, and you can check by a local calculation that for $Y=\mathbb P^2$ this is not the case.
