A step of proof of fundamental theorem of homomorphism of groups

Question

Theorem: Let $G$ be a group. If $N$ is a normal sub group of $G$, then $\frac{G}{N}$ is a homomorphic image of $G$. Conversely, if any group $G'$ is a homomorphic image of $ G$ then $ G'$ is isomorphic to some qoutient group of $G$.

To prove the first part, we define $f:G\rightarrow \frac{G}{N}$ by $f(a)=Na$ for all $a\in G.$ And show that $f$ is an onto homomorphism.

To prove the converse part, define $\phi: \frac{G}{N}\rightarrow G' $ by $\phi(Na)=g(a)$ for all $a\in G.$

Here, why do we need to prove that $\phi$ is a well defined? And why don't we need to show that $f$ is also well defined?

Then we proceed to show that $\phi$ is a one-one and onto homomorphism.

Answer 1

Here, why do we need to prove that $\phi$ is a well defined?

Because the same coset can have different representatives. It is, a priori, possible that $Na=Nb$ but $g(a)\neq g(b)$, which would make $\phi$ not a function. So we must check that this isn't the case.

And why don't we need to show that $f$ is also well defined?

Because if two elements of $G$ are equal, they necessarily give the same coset. So $f$ clearly fulfills the requirements to be a function without any need to actually check.