I want to show that the sheaves $\mathcal{O}$ (holomorphic functions) and $\Omega$ (holomorphic $1$-forms) are isomorphic.
Because every $\omega \in \Omega$ has the form $\omega=f \text{d} z$ for some $f \in \mathcal{O}$ I thought it would be appropriate to construct an isomorphism of sheaves
$$\varphi\colon\mathcal{O} \longrightarrow \Omega \colon f \longmapsto f\text{d}z$$
to show that $\mathcal{O}\cong\Omega$.
Now I want to show that $\varphi$ is indeed an isomorphism of sheaves. How can I show that the diagram is commutative?
I think that $\omega|_U = f|_U \text{d}z$ which would explain the diagram.
I am assuming these sheaves are over $\mathbb{C}$ because $dz$ isn't globally defined in general. If they are, the diagram is trivial as you indicated, restricting from $U$ to $V$ works the same way on $\omega$ and $f$. If this is over $\mathbb{C}^n$ the map should be $f\to f\,dz_1\dots dz_n$.