[Math] Tricks – Prove Homomorphism Maps Identity to Identity – Fraleigh p. 128 Theorem 13.12(1.)

Question

Let $\phi$ be a homomorphism of a group G into a group G'.
If $e =$ the identity element in G, then $\phi(e) =$ the identity element in G'.

Is this what Sharkos is trying to answer:
About homomorphisms, you only know $\phi(a) \phi(b) = \phi(ab)$.
ab is more complicated than anything we want to think about, hence just presuppose $b = e$.
Then $ \begin{align} \phi(a)\phi(e) & = \phi(a\color{magenta}{e}) \\ & = \phi(a) \end{align} $
Left multiply the last equation by $\color{green}{\phi(a)^{-1}}$: $\quad \phi(e) = id_{G'}.$

(1.) How do you predestine to rewrite $a$ as $ a = a\color{magenta}{e}$?
Or to 'presuppose $b = e$' ? I understand neither tricks.

(2.) What's the intuition?

Answer 1

Since $$\phi(e)=\phi(ee)=\phi(e)\phi(e),$$ you can cancel one of the $\phi(e)$ in $G'$, then leaving you with $e'=\phi(e)$.