combinatorical embedding I have a problem with the following statement :

Every combinatorial embedding is equivalent to one with $\lambda(T) = 1$ on a spanning tree of G

What does this mean ? 

OK in a spanning tree there isn't a cycle, so the signature of every edge is equal to $1$ but what is the sense of this statement?
A combinatorial embedding or signed rotation system is a combinatorial representation of a cellular embedding of a graph on some surface.  An embedding is a drawing of the graph such that no two edges cross; an embedding is cellular if every face is a disk.

Consider an undirected graph $G=(V,E)$.  It is helpful to regard each edge in $E$ as a pair of directed edges, or darts.  Each dart goes from one vertex, called its tail, to another vertex, called its head.  The reversal of a dart is obtained by swapping its head and its tail.  Let $\vec{E}$ denote the set of all darts.

A signed rotation system consists of two functions:


A successor permutation $\textit{succ} \colon \vec{E} \to \vec{E}$.  Specifically, $\textit{succ}(\vec{e})$ is the next incoming dart after $\vec{e}$ in "clockwise" order around $\textit{head}(\vec{e})$.  Thus, the cycles of $\textit{succ}$ correspond precisely to the vertices of $G$.
A sign function $\textit{sign}\colon E \to \{-1, +1\}$, which indicates for each edge whether the "clockwise" orders at the endpoints of that edge are consistent ($+1$) or inconsistent ($-1$). 


This representation is not unique.  For example, reversing the cyclic order of incoming darts at any vertex $v$ and changing the signs of all edges incident to $v$ leaves the embedding intact.

The theorem you are asking about is the following:


  Any cellular embedding of any graph $G$ can be encoded by a signed rotation system, such that for some spanning tree $T$, every edge in $T$ is positive.


In fact, one can prove a stronger statement:


  For any spanning tree $T$ of any graph $G$, any cellular embedding of $G$ can be encoded by a signed rotation system, such that every edge in $T$ is positive.


If the underlying surface is orientable, we can choose a global "clockwise" orientation, which means any embedding onto that surface has a representation where all edges in $G$ are positive.

On the other hand, if the underlying surface is non-orientable, some edges must be negative.  Any one-sided cycle must include an odd number of negative edges (and therefore at least one negative edge), because moving once around a one-sided cycle reverses your local orientation.  That's what "one-sided" means.  Still, it is possible to ensure that every edge in your favorite spanning tree is positive.

The proof is straightforward induction over the tree $T$.