Minimal good cover of the torus

Question

Recall that an open cover $\mathfrak{U} = \{ U_\alpha \}$ of a manifold $M$ is called a good cover if all possible finite intersections $U_{\alpha_1} \cap ... \cap U_{\alpha_n}$ are contractible.

Question: What is the minimum number of open sets required for a good cover of the 2-dimensional torus?

The picture below provides a good cover of the torus (i.e. opposite sides of the parallelogram identified as usual) using 7 open sets (i.e. take sufficiently small open neighbourhoods of the hexagons). Can one do any better than 7? If not, how does one prove that 7 is optimal?

Answer 1

You can't do any better than $7$. This follows from

Karoubi, Max; Weibel, Charles A., On the covering type of a space, Enseign. Math. (2) 62, No. 3-4, 457-474 (2016). ZBL1378.55002.

in particular Theorem 5.3 in the arXiv version.

The strict covering type of a space $X$ is the minimal cardinality of a good cover, denoted $\operatorname{sct}(X)$. This is not a homotopy invariant, so Karoubi and Weibel introduce the covering type, defined by $$ \operatorname{ct}(X)=\min\{\operatorname{sct}(X')\mid X'\simeq X\} $$ Obviously $\operatorname{ct}(X)\leq \operatorname{sct}(X)$. In Theorem 5.3 they use cohomological arguments to show that $\operatorname{ct}(T^2)=7$.