Let $p:X \to S$ and $q:Y\to S$ be two objects in the category of ringed spaces over the ringed space $S$, and let $f:X \to Y$ be a morphism over $S$.

Given a sheaf $\mathcal{F}$ of $\mathcal{O}_Y$-modules, there are at least two different ways to produce a morphism $$ q_*\mathcal{F} \to p_*f^*\mathcal{F} $$ of sheaves of $\mathcal{O}_S$-modules by just using canonical operations. We could for instance apply the adjunction unit for $(f^*, f_*)$ to $\mathcal{F}$, push the map forward to $S$ and then use the natural isomorphism $q_*f_* \simeq p_*$. That is $$ \mathcal{F} \to f_*f^*\mathcal{F} $$ $$ q_*\mathcal{F} \to q_*f_*f^*\mathcal{F} $$ $$ q_*\mathcal{F} \to p_*f^*\mathcal{F}. $$ On the other hand, we could start by applying the adjunction co-unit for $(q^*, q_*)$, pull it back to $X$, use the natural isomorpism $f^*q^* \simeq p^*$ and finally use the adjuncton $(p^*, p_*)$ to produce a morphism of $\mathcal{O}_S$-modules. That is $$ q^*q_*\mathcal{F} \to \mathcal{F} $$ $$ f^*q^*q_*\mathcal{F} \to f^*\mathcal{F} $$ $$ p^*q_*\mathcal{F} \to f^*\mathcal{F} $$ $$ q_*\mathcal{F} \to p_*f^*\mathcal{F}. $$

In this case we get the same map (something which at least I find tedious to check; but I might be thinking of it in the wrong way).

My question is:

Is there a general coherence result which frees us from checking such equalities case by case, just as in the case of for instance symmetric monoidal categories. I'm thinking of something like: Start with a commutative diagram of ringed spaces and a map of sheaves of modules over one of the spaces. Is any map produced from this map, by just applying push-forwards and pull-backs and using adjunctions, uniquely determined by the strings of symbols in the domain and co-domain of the new map?

(One could probably state a similar question where we throw the tensor product into the mix of canonical operations.)