From Dr. Pinter's "A Book of Abstract Algebra":
Let $G$ be any group. If $\epsilon: G \rightarrow G$ is the identity function, $\epsilon(x)=x$, show that $\epsilon$ is an isomorphism.
Here's my attempt:
Claim: $\epsilon$ is injective.
Proof: since $\epsilon$ is the identity function, then each element of $B$, i.e. the range, is the image of no more than one element of $A$, i.e. the domain.
Claim: $\epsilon$ is surjective
Proof: since $\epsilon$ is the identity function, then each element of $B$, i.e. the range, will equal its input, i.e. $x$ from $\epsilon(x)$. As a result, each element of the range of at least one element of the domain.
Claim: $\epsilon(ab)=\epsilon(a)\cdot\epsilon(b)$.
Proof: I'm assuming that I can choose multiplication as the operation between $a$ and $b$. As a result, the commutativity of multiplication results completes the proof.
Did I prove that $\epsilon$ is isomorphic?
Well, you can't exactly "choose" what operation $G$ has. It's part of the pack "$G$ is a group", which is actually the short form for "$G$ is a group with operation $\bot$, and we write $\bot(a,b)$ as $ab$". I've used a generic symbol but it could have been any other, this depends on the context.
Anyways, the correct way of concluding is to actually take the image of $ab$ under $\epsilon$ (as we probably would, were $\epsilon$ any other map): $$\epsilon(ab) = ab$$
Now how can you rewrite both $a$ and $b$? The answer is that, since $\epsilon$ is the identity, $a=\epsilon(a)$ and $b = \epsilon(b)$. Therefore the last expression is equal to (by transitivity of the equals sign) $\epsilon(a)\epsilon(b)$.