Problem in defining a map Suppose I have two algebraic objects $\mathcal{A}, \mathcal{B}$ , and I want to define a map between $\mathcal{A}, \mathcal{B}$. So what is a well-defined map and why it has to be well-defined(may be it's a stupid question), and what will happened if it is not well-defined? I read about this in Tim Gowers blog, I got the point however I could not make it clear in algebraic case.
I also have another question. Suppose that $R$ and $S$ are two rings, $\varphi : R \longrightarrow S$ is a ring homomorphism, $\mathfrak{p} \in $Spec$R$, let $I$ be the ideal generated by $\varphi(\mathfrak{p})$ in $S$, $U:=\lbrace\varphi(r)+I|r\in R-\mathfrak{p}\rbrace $. Then we can form the ring $S_{[\mathfrak{p}]}:=U^{-1}(S/I)$. Can we defined a map from $S$ to $S_{[\mathfrak{p}]}$ through two maps $\pi : S\longrightarrow  S/I$ and $f:S/I \longrightarrow U^{-1}(S/I)$? I think we can make a map from $S\times (R-\mathfrak{p})$ to $S_{[\mathfrak{p}]}$. From this, I have to think again about the well-defined property of it, though I still do not know what exactly it means.
I know that my question is not well-written, since my idea in my head still complicated, I beg your pardon for this. 
Thanks for reading and please feel freely commenting and answering my question.
 A: May be the following example will help:
Let $S$ and $T$ be sets, and $f:S \rightarrow T$ be a function. One can define an equivalence relation $\sim$ on  $S$ as follows: for all $x,y \in S$ we define $x \sim y$ if and only if $f(x) = f(y)$. It is quite easy to verify that $\sim$ is indeed an equivalence relation on $S$. Now look at the set of equivalence classes $S/{\sim}$.
You have a natural map $\pi : S \rightarrow S/{\sim}$ where for all $x \in S$, $\pi(x) = [x]_{\sim}$ which is the equivalence class of $x$ in $S/{\sim}$. 
You can also define an map $f' : S/{\sim} \rightarrow T$ as follows: take an arbitrary equivalence class say $A \in S/{\sim}$, and then choose a representative element $x \in A$. Define $f'(A) = f'([x]_{\sim}) = f(x)$. Now from the definition of this map $f'$, it seems that $f'$ really depends on the choice of the representative of the equivalence class $A$. What I mean is, I chose to represent my equivalence class with the particular representative $x$, and my output seems to depend on $x$. But, we know that an equivalence class can have many different representatives. So, instead of choosing $x$, if I choose say $y \in A$, then I would have $[x]_{\sim} = A = [y]_{\sim}$. So, from the way I defined the function $f'$, choosing $y$ as a representative of $A$, I would get $f'(A) = f'([y]_{\sim}) = f(y)$. Now, $f'$ is a function iff $\forall A \in S/{\sim}$, the output of $f'(A)$ is independent of whichever representative I choose for the equivalence class $A$. So, you need to check that the map $f'$ is independent of your choice of a representative of an equivalence class. Another way of saying this is that you need to check that your map $f'$ is well-defined. You probably know this already, but in case you do not, it is a good idea now to check that the map $f'$ is well-defined in the sense I have described.
If Tim Gowers has tried to explain what it means for a function to be well-defined, I doubt I can do any better. But, I hope this helps.
