Let $f: X \to Y$ be a continuous map and $\mathcal{F}$ a soft sheaf over $X$ (let's assume it is a sheaf of sets). I'm trying to prove that $f_{*} \mathcal{F}$ is also soft. (This is stated without proof in page 25 of Grauert's and Remmert's "Theory of Stein Spaces")
So, take the inclusion $i : Z \to Y$ of a closed set. Since $\mathcal{F}$ is soft, we have a global surjective map $\mathcal{F}(X) \to j_{*} j^{-1} \mathcal{F}(X)$, where $j: M \to X$ is also the inclusion of the corresponding closed set, $M = f^{-1}(Z)$. This induces another global surjection $f_{*}\mathcal{F}(Y) \to f_{*}j_{*}j^{-1}\mathcal{F}(Y)$. Now, we also have $$f_{*}j_{*}j^{-1}\mathcal{F} = (f \circ j)_{*} j^{-1} \mathcal{F} = (i \circ f|_{M})_{*} j^{-1} \mathcal{F} = i_{*} (f|_{M})_{*} \mathcal{F}|_{M}$$
I may be wrong, but I believe that we have a canonical iso $(f|_{M})_{*} \mathcal{F}|_{M} \cong (f_{*}\mathcal{F})|_{Z}$. And this would solve it since, then we would have that the canonical adjunction map $f_{*}\mathcal{F}(Y) \to i_{*} i^{-1} f_{*} \mathcal{F}(Y)$ would be surjective, which is what is intented. However, I can't seem to write the isomorphism and I keep getting confused with the sections, so it got very frustrating.
My question then is: Do we have the desired isomorphism? And if so, what is it explicitely?
or, more importantly
Does this hold in full generality or do we need to impose some conditions on $X$, $Y$ or $f$?
I would also be happy if a reference that fully approaches this question is provided.
Thank you for all the help in advance :)