Is this hypothesis of a group homomorphism theorem actually a corollary? I'm trying to finally understand Noether's isomorphism theorems in the context of groups, mainly following Hungerford's treatment. 
The question is regarding his Theorem 5.6 of Chapter 1, which states:

If $f: G \rightarrow H$ is a homomorphism of groups and $N$ is a normal subgoup of $G$ contained in the kernel of $f$, then there is a unique homomorphism $\bar{f}: G/N \rightarrow H$ such that $\bar{f}(aN) \rightarrow f(a)$ for all $a \in G$. $\text{Im} f = \text{Im} \bar{f}$  and $\ker \bar{f} = \ker f / N$. $\bar{f}$ is an isomorphism if and only if $f$ is an epimorphism and $N = \ker f$.  

After proving this on my own, I am not sure why $N \subset \ker f$ is a hypothesis and not a corollary. Please allow me to explain: 
First, I am unable to interpret the claim regarding $\bar{f}$ being the unique homomorphism such that $\bar{f}(aN) \rightarrow f(a)$ as anything but a construction of $\bar{f}$. Furthermore, this means $\bar{f}$ can be regarded as a subset of $G/N \times H$, specifically the subset defined by $\{(aN, f(a)) \ | \ a \in G\}$, which is obviously well-defined in the sense of being single-valued, since $f$ is a homomorphism. That makes the terminology unique strange to me, since apparently $\bar{f}$ is unique by construction. 
Now we must prove $\bar{f}$ is a homomorphism from $G/N$ to $H$. To show  this, take $aN$ and $bN$ from $G/N$. We have
$$
\bar{f}(aNbN) = \bar{f}(abN) = f(ab) = f(a)f(b) = \bar{f}(aN)\bar{f}(bN).
$$ 
If this style of manipulating cosets is unfamiliar, consider Van der Waerden's definition of a complex, which is simply an arbitrary subset of the group. If $A$ and $B$ are complexes then the product $AB$ is defined as $\{ab \ | \ a\in A, b\in B\}.$ Van der Waerden asks which properties $A$ and $B$ must satisfy if $AB$ is a subgroup and concludes $AB < G$ iff $AB = BA$. Writing $aB$ for $\{a\}B$, we are granted an interpretation of the set of normal subgroups of $G$ as the center of the free group over $P(G)$. Thus the coset product $aNbN$ may be regarded as the product of complexes $\{a\}N\{b\}N = \{a\}(N\{b\})N =\{a\}(\{b\}N)N =\{a\}\{b\}N = abN.$ 
Evidently the fact that $\bar{f}$ is a unique homomorphism constructed as given does not rely on $N \subset \ker f$ as a hypothesis. The latter can however be arrived at as a corollary to $\bar{f}$ being a homomorphism. Just compare $\bar{f}(N) = \bar{f}(1_GN)= f(1_G) = 1_H$ with $\bar{f}(N) = \bar{f}(nN) = f(n)$ whenever $n \in N$. 
To conclude, I find that the hypothesis $N \subset \ker f$ is not necessary to conclude $\bar{f}$ is a homomorphism, but follows easily from the fact that it is. I feel like there is something I am missing however. 
 A: If $N$ is not contained in the kernel of $f$, then you can "define" $\overline{f}(aN) = f(a)$, but this is not in fact well defined.
To see this, let $b$ be an arbitrary element of $aN$, so $bN = aN$. Then if $\overline{f}$ is to be well defined we must have $f(b) = f(a)$. This in turn is true if and only if $f(b)f(a^{-1}) = f(a)f(a^{-1})$. Since $f$ is a homomorphism, the latter condition simpifies to $f(ba^{-1}) = f(1) = 1$, or equivalently $ba^{-1} \in K$, where $K$ is the kernel of $f$. We can also write this as $b \in aK$.
Summarizing, we showed that if $\overline{f}$ is well defined, and $b \in aN$, then $b \in aK$. Equivalently, if $\overline{f}$ is well defined, then $aN \subset aK$, which is equivalent to $N \subset K$.
A: The hypothesis $N\subset\ker f$ is necessary to conclude $\overline{f}$ is a well-defined map. Example:  Let $$f:\mathbb{Z}\to\mathbb{Z}_2$$ be the canonical homomorphism. Obviously, $\ker f=2\mathbb{Z}$ and $N=3\mathbb{Z}\not\subset \ker f$. Now we have $1+\mathbb{Z}=4+\mathbb{Z}$ on $\mathbb{Z}/N$ but $1=\overline{f}(1+N)\neq\overline{f}(4+N)=0$ on $\mathbb{Z}_2$. Thus $\overline{f}(x+N)=f(x)$ is not well-defined.
