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Let’s say $f$ maps $R$-module $A$ to $B$, and define $f$ by $f(x) = h(m)$, say, for $x$ in $A$, and $h(m)$ in $B$. The way I know to show this is well-defined is by picking $x$ and $y$ in $A$, where $x = y$, and showing that $f(x) = f(y)$. However, I saw in a snake-lemma proof that the author simply proved the function is well-defined by picking $x$ in $A$, and showing that $f(x)$ is in $B$. So, I want to know if these approaches mean the same? I would be grateful if you could help me shed light on this.

Note: I think if the approaches mean the same, it is because $x$ is arbitrary. However, I am not totally convinced.

Thanks in advance.

citadel
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Mr Prof
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1 Answers1

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As per my knowledge, "well-defined" may have two meanings.

First, when the membership of the image element to the codomain needs to be proven. For example, for $G$ a group, if your candidate map $G\to \operatorname{Sym}(G)$ is defined by $a\mapsto(g\mapsto ag)$, you firstly need to prove that the map $g\mapsto ag$ is indeed a bijection on $G$ (namely an element of $\operatorname{Sym}(G)$). Can you, in this case?

Second, when you are trying to define a map where the domain is a quotient set, and hence you have no way to avoid relying on a (arbitrary) class representative to define the map, you firstly need to prove that such a (arbitrary) choice doesn't affect the result, namely it doesn't change the image element. For example, given a map $f\colon X\to Y$, setting $x_1\sim x_2\stackrel{(def.)}{\iff}f(x_1)=f(x_2)$ defines an equivalence relation on $X$, and the map $\tilde f\colon (X/\sim)\to Y$ defined by $[x]\mapsto f(x)$ is well-defined, because for $[x']=[x]$ you get: $\tilde f([x'])=$ $f(x')\stackrel{x'\in[x]}{=}$ $f(x)=$ $\tilde f([x])$. On the other hand, another map like $[a]\mapsto a$ is clearly not well-defined, as for $a\ne b\in [a]$, you get $[a]=[b]\mapsto b\ne a$.

citadel
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