# What is “well-defined”?

It is very often the case that you may be asked to show a map is well-defined. What does this even mean? Well, it boils down to two main things. Let $f:A\to B$ be some map between sets $A$ and $B$. To check that $f$ is well-defined, we need to ask ourselves the following.

• Is $\im(f)\subset B$ ? If not, then $f$ is not well-defined.
• Is $A$ a collection of sets (e.g. a set of cosets)? If yes, then we must ask, is $f$ defined in terms of choices of representatives of these sets? If so, then we must verify that for all $a_1,a_2\in A$ with $a_1=a_2$, we have $f(a_1)=f(a_2)$. If this does not hold, then $f$ is not well-defined.
If it is the case that $f$ satisfies both of the aforementioned properties, then $f$ is well-defined. Let’s look at some examples to make this clear. Consider the map $f:\reals\to\reals_{<0}$ defined by $f(x)=|x|$. Clearly $\im(f)\not\subset\reals_{<0}$, since the absolute value of any real number is nonnegative. Thus $f$ is not well-defined. On the other hand, the map $f:\reals\to\reals_{\geq0}$ defined by $f(x)=x^2$ is well defined, since it is clear that $\im(f)\subset\reals_{\geq0}$.

Let’s now turn to some more complicated examples.

Let $U,W$ be vector spaces and define the map $f:U\to(U+W)/W$ via $f(u)=u+W$. We need to check that $\im(f)\subset(U+W)/W$. How do we do this? Well, let $u\in U$. Then $f(u)=u+W=u+0+W$, and since $0\in W$, we have $f(u)\in(U+W)/W$. Thus $\im(f)\subset(U+W)/W$, so $f$ is well-defined.

Let $G,H$ be groups and let $N$ be a normal subgroup of $G$. Suppose we have a homomorphism $\phi:G\to H$ with $N\subset\ker(\phi)$. Then we can define the map $\psi:G/N\to H$ via $\psi(gN)=\phi(g)$. Since we are defining $\psi$ in terms of coset representatives, we need to check that if $g_1N=g_2N$, then $\psi(g_1N)=\psi(g_2N)$. So suppose $g_1N=g_2N$ for some $g_1,g_2\in G$. Then $g_1g_2^{-1}\in N$. Note that $\phi(g_1g_2^{-1})=\phi(g_1)\phi(g_2)^{-1}$, and since $N\subset\ker(\phi)$, we have $\phi(g_1)\phi(g_2)^{-1}=1$, so $\phi(g_1)=\phi(g_2)$. This means $\psi(g_1N)=\psi(g_2N)$, so $\psi$ is well-defined.

In most situations, it will be clear if you need to prove a map is well-defined. Namely, if the map is defined on a collection of sets, we must check that the choice of representative does not affect the output, and if it is not entirely obvious that the image of the map is contained in the codomain, this must be checked as well. Otherwise, well-definedness can usually be brushed off as “clear”.

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