Is taking a Koszul complex of a function a functor?

While reading about Koszul complexes in Bruns and Herzog, I came across the following proposition:

Suppose $L$ and $L'$ are $R$-modules with linear forms $f : L \to R$ and $f' : L' \to R$. Every $R$-homomorphism $\phi : L \to L'$ extends to a homomorphism $\bigwedge \phi : \bigwedge L \to \bigwedge L'$ of $R$-algebras, as discussed above. If $f = f' \circ \phi$, then $\bigwedge \phi$ is a homomorphism of Koszul complexes:

Proposition 1.6.8 With the notation just introduced, if $f = f' \circ \phi$, then $\bigwedge \phi : K_*(f) \to K_*(f')$ is a complex homomorphism.

-- Bruns, Winfried and Jurgen Herzog (2005). Cohen-Macaulay Rings. Page 46. Section 1.6.

My understanding of this proposition is that every $R$-module homomorphism $\phi : L \to L'$ induces a map $$\begin{array}{rcl} \phi : \text{Hom}(L',R) & \to & \text{Hom}(L,R) \\ f' & \mapsto & f' \circ \phi. \end{array}$$ This in turn induces chain maps $\bigwedge \phi : K_*(f' \circ \phi) \to K_*(f')$, for every $R$-linear form $f' : L' \to R$. For curiosity's sake, my question is whether I can do the following. (Bear in mind that I do not even know if the categories will be well-defined, since I am not comfortable with category theory.)

Attempted definition Let $\mathcal{L}$ be the category of $R$-linear forms. Its objects are $\text{Hom}(L,R)$, with morphisms $\phi$ as defined above, for each $R$-module homomorphism $\phi$. Let $\mathcal{K}$ be the category of Koszul complexes $K_*(f)$. Its objects are of the form $$K(L) := \{ K_*(f) : \text{Domain of f is L.\} }$$ Its morphisms are the chain complexes between Koszul complexes.

Question Is taking the Koszul complexes of $R$-module homomorphisms a (contravariant) functor $\mathcal{L} \to \mathcal{K}$?

Attempted proof By definition of Koszul complex, there is a map $$\begin{array}{rcl} K_* : \mathcal{L} & \to & \mathcal{K} \\ \text{Hom}(L,R) & \mapsto & K(L) \\ f & \mapsto & K_*(f), \end{array}$$ defined by taking Koszul complexes. The proposition from Bruns and Herzog then shows that every morphism $\phi \in \text{Mor}(\mathcal{L})$ gives a morphism $\bigwedge \phi \in \text{Mor}(\mathcal{K})$. It reverses the order of the objects in the morphism, so it is contravariant if it is a functor. Identity is certainly preserved. Composition is preserved as well: $$\bigwedge(\phi' \circ \phi) : K_*(f'' \circ \phi' \circ \phi) \xrightarrow{\bigwedge\phi} K_*(f'' \circ \phi') \xrightarrow{\bigwedge\phi'} K_*(f'').$$ This shows that $K_*$ is indeed a functor.