Do the link cobordism induced maps on Khovanov homology for different diagrams commute? I'm trying to determine the validity of a property of the morphisms on Khovanov homology that are induced by oriented link cobordisms. These maps are defined with fixed diagrams for the boundary links, but what changes when we choose different diagrams? Is there a relation between the induced morphisms? More explicitly:
Suppose we have an oriented link cobordism $\Sigma \subset \mathbb{R}^3 \times [0,1]$ from $L_0$ to $L_1$ and that we have a pair of diagrams for each link: $D_0, D_0'$ for $L_0$ and $D_1, D_1'$ for $L_1$. There are induced morphisms (defined on the chain level via surface diagrams) on the Khovanov homology:
$$F : \text{Kh}(D_0) \to \text{Kh}(D_1)$$ $$F' : \text{Kh}(D_0') \to \text{Kh}(D_1')$$
How are these maps related? I'm tempted to say that because there are Reidemeister-induced isomorphisms $\varphi_0 : \text{Kh}(D_0) \to \text{Kh}(D_0')$ and $\varphi_1 : \text{Kh}(D_1) \to \text{Kh}(D_1')$, then the following diagram should commute (up to sign).
$\require{AMScd}$
\begin{CD}
\text{Kh}(D_0) @>{\varphi_0}>> \text{Kh}(D_0') \\ @V{F}VV @VV{F'}V \\ \text{Kh}(D_1) @>{\varphi_1}>> \text{Kh}(D_1')
\end{CD}
I haven't found any resources discussing this, and I'm having a hard time (dis)proving it. All of the maps I've come up with have commuted, and I haven't had much of a breakthrough in proving it. 
Here's what I've considered:
Jacobsson proved that the induced map $F$ is invariant up to an overall multiplication by -1 under ambient isotopy of $\Sigma$ leaving $\partial \Sigma$ setwise fixed. The same can be said for $F'$, or any induced map defined in this manner (i.e. by choosing a generic surface diagram beginning/ending with the given diagrams for the boundary links of $\Sigma$ and composing the Reidemeister/Morse induced chain maps that the surface diagram records). 
In the spirit of Jacobsson's theorem, one could try to argue that the maps $F'\varphi_0$ and $\varphi_1F$ are induced by equivalent cobordisms (i.e. cobordisms that are related by a boundary-preserving isotopy). This would imply that these morphisms are the same (up to sign). But constructing these cobordisms is tricky, as they need to have fairly specific surface diagrams. Moreover, surface diagrams (as I understand them) are defined with respect to a projection $\mathbb{R}^3 \to \mathbb{R}$, which is already not consistent between the surface diagrams that induce $F$ and $F'$.
 A: Your diagram should commute and this is a sketch of an argument for that using Jacobsson's theorem.
The diagrams $D_0$ and $D_0'$ are related by a sequence of Reidemeister moves. Each Reidemeister move can be thought of as a cobordism so combining them gives a cobordism $A$ from $D_0$ to $D_0'$. Similarly, a sequence of Reidemeister moves gives a cobordism $B$ from $D'_1$ to $D_1$. These cobordisms $A$ and $B$ induce the maps $\varphi_0$ and $\varphi_1^{-1}$.
Now you have a new cobordism from $D_0$ to $D_1$ given by the composition $B \circ \Sigma_1 \circ A$ which is inducing the map $\varphi_1^{-1} \circ F' \circ \varphi_0 $. The cobordisms $\Sigma_0$ and $B \circ \Sigma_1 \circ A$ are isotopic relative the boundaries because $A$ and $B$ are cobordisms corresponding to the isotopies relating the different diagrams and $\Sigma_0$, $\Sigma_1$ are representations of the same cobordism for these different diagrams. Then Jacobsson's theorem says that the maps $F$ and $\varphi_1^{-1} \circ F' \circ \varphi_0 $ agree up to a sign. This implies that the diagram you've drawn commutes.
