Assume we have a $3\times 3$ grid with rows and columns being short exact sequences of $C^*$-algebras.

This gives a grid of 6-term exact sequences: 3 "horizontal" sequences and 3 "vertical" sequences, forming a torus-like diagram with 18 groups and 36 maps. Commutativity of 16 out of 18 squares follows from naturality of functors $K_0$, $K_1$ the index map $\delta_1$ and exponential map $\delta_0$. Two squares remain, and it seems, that they are "anti-commutative", but I didn't manage to prove this. These squares are the following.

My conjecture is that both squares are anti-commutative, i.e.

- $\delta_0^{V0}\delta_1^{H2}(g) = -\delta_0^{H0}\delta_1^{V2}(g)$ for any $g\in K_1(A_{22})$,
- $\delta_1^{V0}\delta_0^{H2}(g) = -\delta_1^{H0}\delta_0^{V2}(g)$ for any $g\in K_0(A_{22})$.

One follows from the other by replacing $A_i$ with its suspension $SA_i$.

**Question:** Is it possible to prove the anti-commutativity of squares (2) and (3) in general?

P.S. The notation follows the book "An introduction to $K$-theory for $C^*$-algebras" by Rordam, Larsen and Laustsen. Upper index of $\delta_i$ describes the exact sequence from (1) it corresponds to. E.g. $H2$ refers to the bottom (#2) horizontal sequence on diagram (1).