Do the maps of rings coincide under the condition of reducedness and localization assumption? 
Say $f,g:B\rightarrow A$ are maps of rings such that $A$ is reduced and $i_{\mathfrak p}\circ f=i_{\mathfrak p}\circ g $ for all primes $\mathfrak p\subset A$ where $i_{\mathfrak p}:A\rightarrow A_{\mathfrak p} \rightarrow k(\mathfrak p)$. Here, $k(\mathfrak p)=A_p/pA_p$. 
  Is it true that $f=g$ ?

My attempt:
I tried working with the equaliser of $(f,g)$ and I know that kernels of $f$ and $g$ are radical, but I'm stuck.
 A: Note that $f$ and $g$ are really a red herring here: the question is just whether $f(b)=g(b)$ for each $b\in B$ given that $i_\mathfrak{p}(f(b))=i_\mathfrak{p}(g(b))$ for all $\mathfrak{p}$, and $f(b)$ and $g(b)$ are just some elements of $A$.  So really, the question is whether an element of $A$ is determined by its images under every $i_\mathfrak{p}$, or equivalently whether the map $$i:A\to \prod_\mathfrak{p} k(\mathfrak{p})$$ which is given by $i_\mathfrak{p}$ on each coordinate is an injection.  But this is easy: $i$ is a homomorphism, and its kernel is $$\bigcap_\mathfrak{p}\ker(i_\mathfrak{p})=\bigcap_{\mathfrak{p}}\mathfrak{p}.$$  That's the intersection of all prime ideals in $A$, which is the nilradical of $A$ and hence just $0$ since $A$ is reduced.  Thus $i$ is an injection, so $f=g$.
More generally, the fact that this map $i$ is an injection for a reduced ring is generally a handy fact; for instance, it tells you that every reduced ring is isomorphic to a subring of a product of fields, so any property of fields which is preserved by products and subrings also holds in any reduced ring.
