Let \(X\) be the 2 complex obtained from \(S^1\) with its usual cell structure by attaching two 2 cells by maps of degrees 2 and 3, respectively.
Compute the homology groups of all the subcomplexes \(A \subset X\) and the corresponding quotient complexes \(X/A\).
Show that \(X \simeq S^2\) and that the only subcomplex \(A \subset X\) for which the quotient map \(X \to X/A\) is a homotopy equivalence is the trivial subcomplex, the 0 cell.
Solution
Let \(e^0,e^1,e^2_2,e^2_3\) denote the cells, where the subscripts record the degrees of the two attaching maps. The cellular chain complex of \(X\) is \[ 0\longrightarrow \mathbb Z^2 \xrightarrow{d_2}\mathbb Z \xrightarrow{0}\mathbb Z\longrightarrow0, \qquad d_2(a,b)=2a+3b. \] Hence \[ H_2(X)=\ker d_2=\mathbb Z(3,-2), \qquad H_1(X)=0, \qquad H_0(X)=\mathbb Z. \]
1. The subcomplexes of \(X\) are exactly\[ \{e^0\},\quad S^1=e^0\cup e^1,\quad A_2=S^1\cup e^2_2,\quad A_3=S^1\cup e^2_3,\quad X. \]Their homology groups are\[ \begin{array}{c|ccc} A&H_0&H_1&H_2\\ \hline \{e^0\}&\mathbb Z&0&0\\ S^1&\mathbb Z&\mathbb Z&0\\ A_2&\mathbb Z&\mathbb Z_2&0\\ A_3&\mathbb Z&\mathbb Z_3&0\\ X&\mathbb Z&0&\mathbb Z. \end{array} \]
A subcomplex containing a \(2\)-cell must contain its attaching circle, hence \(e^1\). This gives the displayed list. The chain complexes for \(A_2\) and \(A_3\) are respectively \[ 0\to\mathbb Z\xrightarrow{2}\mathbb Z\to\mathbb Z\to0 \] and \[ 0\to\mathbb Z\xrightarrow{3}\mathbb Z\to\mathbb Z\to0, \] which give the stated groups. The groups for \(X\) follow from the cellular complex above.
2. The quotient spaces and their homology are as follows:\[ X/\{e^0\}=X, \qquad X/S^1\cong S^2\vee S^2, \qquad X/A_2\cong S^2, \qquad X/A_3\cong S^2, \qquad X/X=\mathrm{pt}. \]Thus\[ \widetilde H_2(X/S^1)\cong\mathbb Z^2, \quad \widetilde H_2(X/A_2)\cong\mathbb Z, \quad \widetilde H_2(X/A_3)\cong\mathbb Z, \]and all other reduced homology groups of these three quotients vanish.
Collapsing the \(1\)-skeleton turns each \(2\)-cell into a \(2\)-sphere, giving \(S^2\vee S^2\). If \(A_2\) is collapsed, only the interior of \(e^2_3\) remains, with its entire boundary collapsed, giving \(S^2\); similarly for \(A_3\).
3. The space \(X\) is homotopy equivalent to \(S^2\).
First attach the degree-\(2\) cell, obtaining \(A_2\). Its fundamental group is \[ \pi_1(A_2)\cong\mathbb Z_2. \] The degree-\(3\) attaching loop for the second \(2\)-cell represents the same element of this group as a degree-\(1\) loop, since \(3\equiv1\pmod2\). Hence these two attaching maps are homotopic in \(A_2\). Replacing the degree-\(3\) attaching map by the homotopic degree-\(1\) attaching map does not change the homotopy type of the adjunction space.
With this replacement, \(S^1\cup e^2_3\) is a disk, so the remaining degree-\(2\) cell is attached to a contractible disk along a loop that is nullhomotopic there. Therefore the resulting space is homotopy equivalent to \[ D^2\vee S^2\simeq S^2. \] Thus \(X\simeq S^2\).
4. For \(A=A_2\), the quotient map\[ q_2:X\to X/A_2\cong S^2 \]induces multiplication by \(-2\) on \(H_2\), up to orientation; for \(A=A_3\), the quotient map induces multiplication by \(3\), up to orientation.
The generator of \(H_2(X)\) is the cellular cycle \[ 3e^2_2-2e^2_3. \] After collapsing \(A_2\), the first term disappears and the second \(2\)-cell generates \(H_2(X/A_2)\), so \[ (q_2)_*(3,-2)=-2. \] After collapsing \(A_3\), only the first cell survives, giving \[ (q_3)_*(3,-2)=3. \] Changing orientations changes only the signs.
5. The quotient map \(X\to X/A\) is a homotopy equivalence only for the trivial subcomplex \(A=\{e^0\}\).
For \(A=S^1\), the quotient has \(H_2\cong\mathbb Z^2\), unlike \(H_2(X)\cong\mathbb Z\). For \(A=A_2\) or \(A=A_3\), the quotient is a sphere, but by 4 the induced map on \(H_2\) is multiplication by \(2\) or \(3\), not an isomorphism. For \(A=X\) the quotient is a point. Finally collapsing the single \(0\)-cell does nothing: \(X/\{e^0\}\) is canonically homeomorphic to \(X\). Hence precisely the trivial subcomplex gives a quotient homotopy equivalence.