# Image of normal subgroup under surjective homomorphism as kernel

Let $f: G \to H$ be a group homomorphism. Then the preimage $f^{-1}(N)$ of a normal subgroup $N$ of $H$ is normal in $G$: this preimage is the kernel of the composition of $f$ with the canonical projection $H \to H/N$.

Now, if $K$ is a normal subgroup in $G$, and if additionally $f$ is surjective, then the image $f(K)$ is normal in $H$, which is easy to see 'elementary' (on the level of elements)... but I would like to recognize the image as normal by recognizing it as some kernel (i.e. along the lines of the above argument). How?

We want to push, by some homomorphism, $f(K)$ to identity (and no other elements to identity).
How can we do this: first pull-back $f(K)$ via (inverse of) $f$: sine $\ker(f)=N$, we have $$f^{-1}(f(K))=KN \,\,\,\,(\mbox{which is normal in G})$$ Next, push this $KN$ (and only that) to identity via a homomorphism; the most natural choice is $$\varphi\colon G\rightarrow G/KN.$$
Thus, we define $\eta\colon H\rightarrow G/KN$ by $\eta=\varphi\circ f^{-1}$.
(You may try to show: $\eta$ is well defined and also a homomorphism).
Claim: $f(K)$ is the kernel of this homomorphism $\eta$.
Apply $\eta$ on $f(K)$: $$\eta(f(K))=\varphi\circ f^{-1}(f(K))=\varphi(KN)=\{1\}.$$ So $f(K)\subseteq \ker\eta$. Converlely, $$\eta(x)=1\Longleftrightarrow \varphi(f^{-1}(x))=1\Longleftrightarrow f^{-1}(x)\in \ker\varphi=KN\Longrightarrow x\in f(KN)=f(K).$$