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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.

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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.

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    I think my answer should extend to algebraic objects without any serious modifications, because in algebra you often deal with homomorphisms of quotient groups or quotient rings, whose underlying sets are just equivalence classes of the original group or ring.2012-03-28
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    Also localizations of rings and modules are essentially equivalence classes with additional structure.2012-03-28