Half a year ago, when I heard the concept of **well-defined
functions**, I *wasn’t* familiar with it.

I’ve just worked out a problem, and got more idea about that concept. The problem should be quite easy. It asks readers to show that if $\langle G, * \rangle$ is a group, $g \in G$ and $\varphi_g: G \to G$ is a mapping defined by $\varphi_g (x) = g * x * g^{-1}$, then $\varphi_g: G \to G$ is an automorphism. However, I misunderstood the wordings in the question, and attempted to prove that the binary structure $\langle \{\varphi_g \mid g \in G\},\circ \rangle$ is isomorphic to $\langle G, * \rangle$, where $\circ$ denotes the composition of functions. As a result, I let $\phi: \{\varphi_g \mid g \in G\} \to G$ be a mapping defined by $\phi(\varphi_g) = g$. The associativity, existence of identity element and existence of inverse of the binary structure can be easily verified. By the very definition of $\phi$, it seems that its surjectivity is very obvious. I continued to write “injectivity of $\phi$ is also obvious.”

\[ \phi(\varphi_{g_1}) = \phi(\varphi_{g_2}) \iff g_1 = g_2 \implies \varphi_{g_1} = \varphi_{g_2} \]

I tried to turn the above rightward double arrow ‘$\implies$’ into a
double-headed one. If I *couldn’t* do so, it means that
$\phi(\varphi_{g_1}) = \phi(\varphi_{g_2})$, though $g_1 \neq
g_2$. I realised that I need to check whether $\phi$ was
*well-defined*. As a result, I wasted an hour on some equations.
Suddenly, I stopped substituting $x = g_1$ or $x = g_2$ into
$\varphi_{g_1} (x) = \varphi_{g_2} (x)$. Instead I took $\langle
G, * \rangle = \langle \R, \cdot \rangle$ and realized what I just did
was a waste of time.