Q:

Cell complex of torus

Suppose we have a plane model for torus. We want to decompose the plane model of torus into cell complex which satisfies
1)every face of the cell complex has exactly 6 boundary edges
2)at each vertex, there are 3 faces meet at it.
How to draw such cell complex? I have try to squeeze a set of hexagons which intersect each other along side, but I can't manage to put this set into plane model of torus to satisfy the conditions above. 
Can anyone enlighten me?

A:

I'm not sure what you mean by 'decompose a plane model for a torus' so I'll just outline how to give a CW-complex construction of a torus which satisfies your conditions.
A hexagon with side edge pairings $[abca^{-1}b^{-1}c^{-1}]$ is homeomorphic to the torus (seen by cutting along the line from the center of one $a$ edge to the center of the other $a$ edge and then gluing along the $c$ edges). This doesn't yet satisfy your first condition because the face only has three boundary edges. We can consider a union of fundamental regions though which does satisfy the conditions.
I drew the region in paint shown below, but the side pairing sequence is given as (the letters used refer to the colours in the image e.g $r=\color{red}{red}$, $m=\color{purple}{mauve}$, $p=\color{pink}{pink}$, ...) $$[gbpmyrb^{-1}g^{-1}m^{-1}p^{-1}r^{-1}y^{-1}]$$ which uses three of the above hexagons glued together to give a different fundamental region of a free/transitive $\mathbb{Z}^2$-action on the plane. This better satisfies the conditions you ask for because each vertex shares three different faces, and we have that each face properly has three boundary edges.

A:

Here's a picture of such a cellulation of the torus. Identify opposite edges of the hexagon as in Daniel Rust's answer.

