Difference between blowup of $V_1\cup V_2$ and union of blowup of $V_1$ and blowup of $V_2$ Let $V_1,V_2\subset \mathbb C^N$ be sub varieties and let $C=\{f_1=\dots=f_d=0\}$ be a subvariety of $\mathbb C^N$. Consider the blowup of $\mathbb C^N$ along $C$ and denote it by $B_C\mathbb C^N$. Then, $B_C\mathbb C^N\subset \mathbb C^N\times \mathbb P^{d-1}$ and since blowups respect closed immersions, $$B_C(V_1\cup V_2)\subset B_C\mathbb C^N,\\ B_C(V_1)\subset B_C\mathbb C^N,\\ B_C(V_2)\subset B_C\mathbb C^N,$$ and we may of course form the union $B_C(V_1)\cup B_C(V_2)$.
Question: Is there a difference between $B_C(V_1\cup V_2)$ and $B_C(V_1)\cup B_C(V_2)$?
I tried as basic example the union of the coordinate axis $V_1=\{x=0\}$, $V_2=\{y=0\}$ inside $\mathbb C^2$ and took as center there point of intersection. Here, $B_C(V_1\cup V_2)$ and $B_C(V_1)\cup B_C(V_2)$ turn out to be the same. I searched on the internet and couldn't find a general statement regarding this situation (which usually indicates that it is wrong and that there are simple counterexamples).
I am therefore asking for a confirmation or a counterexample. If the answer depends on the situation and there are criteria to decide this question, I would also like to know them.
If you reference to a good textbook, it's also fine. Thanks.

Yes, it is indeed the same because outside of the point of intersection, as in your example $(0,0)$, you have an isomorphism of algebraic varieties (I'm talking in general without restricting at the field $C$, in your case you have to substitute it with biolomorphism), then from ($V_{1} \setminus (0,0)$)$\cap$($V_{2} \setminus (0,0)$)=$\emptyset$ follows easily that also the intersection of the preimages of the two sets is empty. I don't know your notations, so please let me use $B_{C}(0,0)$ to notate the preimage of $(0,0)$ under the blowup.
So $$B_{C}(V_{1})\cup B_{C}(V_{2}) = B_{C}(V_{1} \setminus (0,0))\cup B_{C}(V_{2} \setminus (0,0))\cup B_{C}(0,0)$$
and
$$B_{C}(V_{1}\cup V_{2}) = B_{C}(V_{1} \setminus (0,0) \cup V_{2} \setminus (0,0))\cup B_{C}(0,0)$$.
The equality $B_{C}(V_{1})\cup B_{C}(V_{2}) = B_{C}(V_{1}\cup V_{2})$ follows from $$B_{C}(V_{1} \setminus (0,0))\cup B_{C}(V_{2} \setminus (0,0)) =  B_{C}(V_{1} \setminus (0,0) \cup V_{2} \setminus (0,0))$$.
