Topology of the torus 
Theorem: There is no open covering $T^k=U_1\cup...\cup U_k$ of the $k$-torus such that the map $$H_1(V,\mathbb{Z}) \rightarrow H_1(T^k,\mathbb{Z}) $$ has rank at most  $(i-1)$ for every component $V$ of $U_i$. 

McMullen states this in his 2004 article on Minkowski's conjecture without proof and without any reference. 
I have no idea how to prove it. Can anyone suggest a proof or give any reference? 
 A: This follows from a more general result of McMullen:
Definition: The order of a cover $\mathfrak{U}$  is the greatest integer $n$ so that  the intersection  of $n+1$ elements is nontrivial.
Loosely stated theorem: Let $\mathfrak{U}$ be an open cover of the $n$-torus. Suppose for that all components $V$ of the intersection $U_1 \cap\dots\cap U_k$ with $k \leq n$, we have that the induced map by inclusion on  homology has at most rank $(n-k)$, then $\mathfrak{U}$ has order at least $n$.
The basic ingredient for the proof is the Cech DeRham complex, reference for which can be found here, on page 6.
A: In Mc Mullen paper "Minkowski's conjecture, well rounded lattices and topological dimension" it is referred as Corollary 2.2, and it is proved at p.5, as a trivial consequence of Theorem 2.1, which is the "Loosely stated theorem" as pointed out by Andres Mejia.
Sincerely, to me, is not clear how that theorem implies the corollary you are looking for, so I cannot help you with that.
A: I think the statement follows from the theorem cited above in the following way:
Suppose there was a cover $(U_i)_{i\leq k}$ such that $$H_1(V)\rightarrow H_1(T^k)$$ has rank at most $i-1$ for each component $V$ of $U_i$. Let $I\subseteq \lbrace 1,\dots,k\rbrace$, $\#I=l$. Then $j=\min I\leq k-(l-1)$. Let $V\subseteq\bigcap_I U_i$ be a component of the corresponding $l$-intersection. Then there is another component $W\subseteq U_j$ with $V\subseteq W$. It follows $$\mathrm{rk}(H_1(V)\rightarrow H_1(T^k))\leq\mathrm{rk}(H_1(W)\rightarrow H_1(T^k))\leq j-1\leq k-l.$$
Therefore, the order of $(U_i)_{i\leq k}$ is at least $k$, which is impossible.
