I am working on the kernels and co-kernels of morphisms in categories with zero objects. I came across the following: $(C,j:B\to C)$ is a cokernel of $f\in \text{Hom}_{\mathcal{C}}(A,B)$ $\iff$ $(C,j^{op}:C\to B)$ is a kernel of $f^{op}\in \text{Hom}_{\mathcal{C}^{\text{op}}}(B,A)$.
Here, the category $\mathcal{C}$ has a zero object and kernel of any morphism $f:A\to B$ in $\mathcal{C}$ is defined as the the pair $(Z,i:Z\to A)$ (for $Z\in\text{Obj}(\mathcal{C})$ and morphism $i\in \text{Hom}(Z,A)$) such that $f\circ i=0_{ZB}$ where $0_{ZB}:Z\to B$ is the zero morphism.
So, I was wondering if you have a commutative diagram in $\mathcal{C}$ and reversed all the arrows to get a diagram in $\mathcal{C}^\text{op}$, then will this diagram also be commutative?
Here are some of my confused thoughts about this. Take $D:\mathcal{C}\to \mathcal{C}^\text{op}$ the functor s.t. $D: A\mapsto A$, $[f:A\to B]\mapsto [f^\text{op}:B\to A]$. This is a contravariant functor becuase the composition $A\to B\to C$ must get mapped to its reverse/dual $C\to B\to A$ in $\mathcal{C}^\text{op}$, by definition of composition in opposite categories (very unsure of this statement). So, going back to my problem about kernels, supposing $(Z,i:Z\to A)$ is kernel of $f:A\to B$ i.e. $f\circ i=0_{ZB}$, we get $i^{op}\circ f^{op}=D(i)\circ D(f)=D(f\circ i)=D(0_{ZB})=0_{BZ})$ i.e. $(Z,i^{op}:A\to Z)$ is the co-kernel of $f^{op}:B\to A$.
I think my confusion is about my understanding of opposite categories and their relation to contravariant functors. For instance, should the composition in $D(i)\circ D(f)$ be that of the opposite category? Is it right to say $f^{op}\circ_{op}g^{op}$ is $D(g\circ f)=(g\circ f)^{op}$?
Any help will be greatly appreciated!!