Image of a superfluous module is superfluous

Question

Let $f$ be a homomorphism between two modules $M$ and $N$.

If $K$ is superfluous in $M$ ($K$ superfluous means that if $K+L=M$ then $L=M$), then $f(K)$ is superfluous in $N$.

The proof I found, but which I dont understand:

Consider some $H$ submodule in $N$, such that $f(K)+H =N$. Then $f^{-1}(H) + K = M$ and so $f^{-1}(H)=M$. Therefore $H$ contains the image of $f$, and in particular in contains $f(K)$, so $H=N$.

What I don't understand is how $f^{-1}(H) + K = M$ follows from $f(K)+H = N$. I only see that $f^{-1}(f(K)) + f^{-1}(H)=M$ .

Answer 1

It's just element chasing. Maybe if I write it this way:

Firstly note that $f(K)+H=N\implies f^{-1}(f(K)+H)=f^{-1}(N)=M $.

Now we claim that $f^{-1}(f(K)+H) =f^{-1}(H)+K$.

The containment $\supseteq$ is obvious.

Now suppose that $x$ is in the left-hand side. We know that $f(x)=f(k)+h$ for some $k\in K$, $h\in H$.

That implies $f(x-k)=h\in H$, so $x-k\in f^{-1}(H)$. Hence $x\in f^{-1}(H)+K$. The containment $\subseteq$ has been proven.