Suppose we have a diagram $\require{AMScd}$ \begin{CD} A @>a>> B\\ @V b V V @VV c V\\ C @>>d> D @>e>> E \\ @VfVV @VVgV @VVhV \\ F @>>i> G @>>j> H \end{CD} of schemes (or could be topological spaces instead), and we have an $\ell$-adic sheaf $\mathcal{K}$ on $H$ (or could be a constructible sheaf on a topological sheaf).

Assume the following squares are (commutative and) Cartesian: $\require{AMScd}$ \begin{CD} A @>a>> B\\ @V b V V @VV c V\\ C @>>d> D \end{CD}\begin{CD} D @>a>> E\\ @V b V V @VV c V\\ G @>>d> H \end{CD}

$\require{AMScd}$ \begin{CD} A @>>d> B\\ @Vf \circ b VV @VVg \circ c V \\ F @>>i> G \end{CD}

$\require{AMScd}$ \begin{CD} C @>>e \circ d> E\\ @Vf \circ b VV @VVg \circ c V \\ F @>>j \circ i> G \end{CD}

but $\require{AMScd}$ \begin{CD} C @>>d> D\\ @VfVV @VVhV \\ F @>>i> G \end{CD} is not even commutative. In any Cartesian square, there is a base change morphism of $*$ and $!$-pullbacks, for example, $a^* c^! \rightarrow b^! d^*$. We consider pulling back $\mathcal{K}$ from $H$ to $A$, using $!$-pullback along the vertical morphisms and $*$-pullback along the horizontal morphisms. This yields two morphisms with source the sheaf obtained by pulling back along the outer right path from $H$ to $A$, and target the sheaf obtained by pulling back along the bottom left path from $H$ to $A$:

$$a^* c^! e^* h^! \mathcal{K} \rightarrow b^! f^! i^* j^* \mathcal{K}$$

1. The first comes from composing the base change maps obtained by cutting up the diagram into two Cartesian squares as below

$\require{AMScd}$ \begin{CD} A @>a>> B\\ @V b V V @VV c V\\ C @>>d> D @>e>> E \\ @VfVV @. @VVhV \\ F @>>i> G @>>j> H \end{CD}

2. The second comes from composing the base change maps obtained by cutting up the diagram into two Cartesian squares as below $\require{AMScd}$ \begin{CD} A @>a>> B\\ @V b V V @VV c V\\ C @. D @>e>> E \\ @VfVV @VVgV @VVhV \\ F @>>i> G @>>j> H \end{CD}

Question: must the two morphisms $a^* c^! e^* h^! \mathcal{K} \rightarrow b^! f^! i^* j^* \mathcal{K}$ agree?

One usually assumes heuristically that two canonical morphisms between the same source and target agree, but in this case I feel unsure. If the non-commutative diagram were commutative (hence automatically Cartesian), then it would be immediate.

Finally, an example of this situation:

$\require{AMScd}$ \begin{CD} A @>a>> B\\ @V b V V @VV c = (pt, c_2) V\\ C @>>d = \Delta> C \times C @>e = pr_1>> C \\ @VfVV @VVg = pr_2 V @VVhV \\ pt @>>i> C @>>j> pt \end{CD}