Let $L : \mathbb{R}^n \twoheadrightarrow \mathbb{R}^m$ be a linear map that is surjective. Let $[0,1]^n$ denote the unit cube in $\mathbb{R}^n$, and let $Z := L([0,1]^n) \hspace{5pt}$ ($Z$ is a zonotope). Let $T : \mathbb{R}^N \to \mathbb{R}^m$ be linear with $T([0,1]^N) \subset Z$.
Is there a linear map $T' : \mathbb{R}^N \to \mathbb{R}^n$ so that $$ L \circ T' = T \hspace{30pt} \text{and} \hspace{30pt} T'([0,1]^N) \subset [0,1]^n \text{ ?} $$
Or in "categorical" terms, is there a linear $T'$ that makes this diagram commute ?
So $T'$ might be called a "lift" of $T$ restricted to the appropriate cube (the 3 maps on the edges of the triangle are restrictions of linear maps to cubes, but the restriction notation is suppressed).
Remark 1. If $T'$ exists, then $T([0,1]^N) \subset Z$ (trivially). So the question asks about the converse of this.
Remark 2. $T([0,1]^N)$ is a subzonotope of $Z$. And $T'([0,1]^N)$ is a zonotope that "covers" $T([0,1]^N)$.
Remark 3. If $[0,1]^n$ is replaced by the Euclidean unit ball $B^n$ (and similarly for $[0,1]^N$) then the answer is yes. We can let $T' := L^+ \circ T$, where $L^+$ is the pseudo-inverse of $L$.
Remark 4. If $m=n$, this is trivial. $L$ is invertible. Take $T' := L^{-1} \circ T$.
Remark 5. If $N=1$, this is trivial. Pick any $v \in [0,1]^n$ so $L(v) = T(1)$. Then define $T'(1) = v$.
Remark 6. If $m=1$, this is relatively easy. Let $e_i$ denote the $i$'th elementary basis vector. If all $L(e_i) > 0, ~ i=1,\ldots,n$, then we can define $T'(e_j) := T(e_j) [ L(\sum e_i) ]^{-1} \sum e_i$. In the general case then a similar formula works, depending on the sign of $T(e_j)$.
Remark 7. The simplest non-straightforward case is $n=3$, $m=2$, and $N=2$. I think the answer is yes, but cannot prove it or find a counterexample. A few days ago I submitted this case as separate question: A property of a linear image of the cube
Thank you.