Support of a section behaves well with quotients Let $A$ be reduced Noetherian, $f \in A$, $\mathcal{q}$ a prime ideal of $A$, and $\overline{f} = f + \mathcal{q}$ the image in the quotient. Is it true that
$$
\operatorname{Supp}(\overline{f}) = \operatorname{Supp}(f) \cap V(\mathcal{q})
$$
as subsets of $V(q) \cong \operatorname{Spec}(A/\mathcal{q})$? Or what are the conditions necessary for this equality to hold? 
Edit: This question was a useful step towards my proof that a reduced scheme has no embedded points. This result, combined with the fact that an integral domain has no embedded points, immediately implies the stronger and more familiar result
$$
   \operatorname{Supp}(f) \cap V(q) = V(q)
$$
since the quotient $A/q$ is an integral domain.
 A: I was able to prove this result while writing up the question, and don't see these basic properties of support stated easily visible on the internet, so I thought I would write up my solution for feedback.
The key property is that $\operatorname{Supp} f = \overline{D(f)}$. One containment follows from the fact that if $f \not\in p$, then $f$ becomes a unit, and thus cannot be zero, in the localization at $p$. Hence $\operatorname{Supp} f \supseteq D(f)$, and since the support of a section of a sheaf is always a closed set, $\operatorname{Supp} f \supseteq \overline{D(f)}$.
For the opposite containment, we show that for any $p \in \operatorname{Supp} f$, there is some generization $p' \subseteq p$ which does not contain $f$ (and so is in $D(f)$). For a proof of this fact, look in the localization $A_p$ (which is reduced) and note that $f$ is in all prime ideals if and only if it is zero. 
I think this is an interesting geometric property: the "typical" way for $f \neq 0 \in A_p$ is for $f$ to localize to a unit, but in the case that $p$ is a boundary point with $f \neq 0 \in A_p$ and $f \in p$, then there is some further localization corresponding to larger irreducible closed set containing $\overline{p}$ on which $f$ is generically invertible.
Once we know $\operatorname{Supp} f = \overline{D(f)}$, then applying this both to $f$ and $\overline{f}$ gives
$$
    \operatorname{Supp} \overline{f} = \overline{D(\overline{f})} = \overline{D(f)} \cap \operatorname{Spec} A/p_i = \operatorname{Supp} f \cap \operatorname{Spec} A/p_i.
$$
