Let $\mathcal A$ be a monoidal category. We know that $\mathcal A$ is monoidally equivalent to a strict monoidal category $\mathcal A^{\mathrm{str}}$. In many books/papers it is assumed without loss of generality that $\mathcal A$ is strict to make things easier and not have to worry about associator and unitor isomorphisms. But how do I know that things I prove about $\mathcal A$ in the strict case will also hold if $\mathcal A$ is not strict?

Here is an explicit example of what I am talking about. Assume $\mathcal A$ is a braided monoidal category and suppose $(C, \Delta, \epsilon), (C', \Delta' , \epsilon')$ are coalgebras in $\mathcal A$. This means $\Delta \colon C \to C \otimes C$ is a morphism and $\epsilon \colon C \to I$ is a morphism such that $(1 \otimes \Delta)\Delta = \alpha_{C,C,C}(\Delta \otimes 1)\Delta$ and $1_C = \rho_C (1 \otimes \epsilon)\Delta = \lambda_C(\epsilon \otimes 1) \Delta$ (and similarly for $C'$). I want to show that $C \otimes C'$ becomes a coalgebra. It does, and the comultiplication $\tilde \Delta$ is given by $$ C \otimes C' \xrightarrow{\Delta \otimes \Delta'} (C \otimes C) \otimes (C' \otimes C') \xrightarrow{\alpha} C \otimes (C \otimes (C' \otimes C') \xrightarrow{1 \otimes \alpha} C \otimes ((C \otimes C') \otimes C') \xrightarrow{1 \otimes(\sigma_{C,C} \otimes 1)} C \otimes ((C' \otimes C) \otimes C') \xrightarrow{1 \otimes \alpha^{-1}} C \otimes ( C' \otimes (C \otimes C'))\xrightarrow{\alpha^{-1} } (C \otimes C) \otimes (C' \otimes C') $$ However, this is absolutely terrible and checking that it does give a comultiplication is excruciating. In the strict case it is much easier since all the $\alpha$'s will disappear. How can I be sure (in general) that if a diagram commutes in the strict case then it will commute in the non-strict case with all the extra isomorphisms?