Identify (with proof, but of course you can appeal to the classification of surfaces) all of the compact surfaces without boundary that have a cell decomposition having exactly one 0-cell and exactly two 1-cells (with no restriction on the number of cells of dimension larger than 1).
Solution
1. Any such cell decomposition is connected and has a finite number, say \(r\), of \(2\)-cells.
The unique \(0\)-cell lies in the \(0\)-skeleton. Each \(1\)-cell is attached to that vertex at both endpoints, so the \(1\)-skeleton is connected. Every \(2\)-cell is attached to the \(1\)-skeleton, hence the whole surface is connected.
Since the surface is compact and the given cell decomposition is a CW decomposition, the surface is contained in a finite subcomplex; because the whole space is the surface, only finitely many cells occur. As the ambient space is a surface, the cells have dimensions at most \(2\). Let \(r\) denote the number of \(2\)-cells.
2. One necessarily has\[ r\ge1. \]
If \(r=0\), the surface would equal its \(1\)-skeleton, a graph with one vertex and two edges. Its second homology with \(\mathbb Z/2\) coefficients would therefore vanish.
On the other hand, every connected compact surface without boundary has a mod-\(2\) fundamental class, so \[ H_2(S;\mathbb Z/2)\cong\mathbb Z/2. \] This contradiction shows that at least one \(2\)-cell is present.
3. The Euler characteristic satisfies\[ \chi(S)=1-2+r=r-1\ge0. \]
For a finite CW decomposition, \[ \chi(S)=c_0-c_1+c_2. \] Here \[ c_0=1, \qquad c_1=2, \qquad c_2=r. \] Thus \[ \chi(S)=1-2+r=r-1. \] By 2, \(r\ge1\), so \(\chi(S)\ge0\).
4. By the classification of compact connected surfaces, the only possible surfaces are\[ S^2, \qquad \RP^2, \qquad T^2, \qquad K, \]where \(K\) is the Klein bottle.
For the closed orientable surface \(\Sigma_g\) of genus \(g\), \[ \chi(\Sigma_g)=2-2g. \] The condition \(\chi\ge0\) gives \[ g=0\quad\text{or}\quad g=1, \] namely \(S^2\) and \(T^2\).
For the closed nonorientable surface \(N_k\) of nonorientable genus \(k\ge1\), \[ \chi(N_k)=2-k. \] The condition \(\chi\ge0\) gives \[ k=1\quad\text{or}\quad k=2, \] namely \(\RP^2\) and the Klein bottle. No other closed connected surface has nonnegative Euler characteristic.
5. The sphere \(S^2\) admits a decomposition with one \(0\)-cell and two \(1\)-cells.
Embed a figure-eight graph \[ S^1\vee S^1 \] in \(S^2\) so that the two circles meet at a single point. Use the common point as the unique \(0\)-cell and the two circle interiors as the two \(1\)-cells. The complement of this figure-eight in \(S^2\) has three connected components, each an open disk. Taking these three components as \(2\)-cells gives a CW decomposition with cell counts \[ (c_0,c_1,c_2)=(1,2,3). \] Indeed, its Euler characteristic is \[ 1-2+3=2=\chi(S^2). \]
6. The projective plane \(\RP^2\) admits such a decomposition with two \(2\)-cells.
Start with the standard polygon model of \(\RP^2\) as a \(2\)-gon whose boundary word is \[ aa. \] Its two polygon vertices are identified to one vertex, and the two boundary sides are identified to one \(1\)-cell \(a\).
Inside the polygon add an arc \(b\) joining the two vertices. After the vertex identification, the interior of this arc is a second \(1\)-cell whose two endpoints are the unique vertex. The arc divides the polygon into two disks, which become two \(2\)-cells after the quotient. This is merely a subdivision of the standard polygon model, so the quotient is still \(\RP^2\). Thus \[ (c_0,c_1,c_2)=(1,2,2). \]
7. The torus \(T^2\) admits such a decomposition with one \(2\)-cell.
Use the standard square model with boundary word \[ aba^{-1}b^{-1}. \] All four vertices are identified to one \(0\)-cell, the edge classes \(a\) and \(b\) give exactly two \(1\)-cells, and the square interior is one \(2\)-cell. Hence \[ (c_0,c_1,c_2)=(1,2,1). \]
8. The Klein bottle \(K\) also admits such a decomposition with one \(2\)-cell.
Use the standard square model of the Klein bottle with boundary word \[ aba^{-1}b. \] Again all four vertices become one \(0\)-cell, the two edge classes \(a,b\) give exactly two \(1\)-cells, and the square interior gives one \(2\)-cell. Thus \[ (c_0,c_1,c_2)=(1,2,1). \]
9. Therefore the complete list is\[ \boxed{S^2,\ \RP^2,\ T^2,\ \text{and the Klein bottle}.} \]
Necessity follows from <1>3–<1>4, and <1>5–<1>8 explicitly realize every surface on the list.