Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 20F (Algebraic Topology)

The geometric realization of a simplicial complex consists of the filled triangle $⟨v_1,v_2,v_3⟩$ together with the two extra edges $⟨v_1,v_4⟩$ and $⟨v_2,v_4⟩$. Thus the triangle on $v_1,v_2,v_4$ has only its boundary:

Use the orientationsThe simplicial chain complex haswithThe map ${\partial }_2$ is injective, so $H_2(K)=0$. The complex is connected, so $H_0(K)\cong \mathbb{Z}$ and $\mathrm{rank}{\partial }_1=3$. Hence $\ker {\partial }_1$ has rank $5-3=2$; quotienting by the rank-one primitive subgroup $\mathrm{im}{\partial }_2$ givesThere are no higher simplices. ThereforeThe result agrees with the Euler characteristic $4-5+1=0$.

Define a simplicial map $j:K\to L$ on vertices byThe image of every simplex is a simplex of $L$, and $j\circ i={\mathrm{id}}_L$. The maps $i\circ j$ and ${\mathrm{id}}_K$ are contiguous simplicial maps, since the union of their images on every simplex is contained in a simplex of $K$. Their realizations are therefore homotopic, so $|j|$ is a homotopy inverse of $|i|$.

Here is an explicit chain homotopy. Put $f=i_{\bullet }{j}_{\bullet }$ and use the orientations from part (b). DefineWe verifyIt is immediate on every vertex except $v_3$, where $\partial h_0(v_3)=v_3-v_1=v_3-f(v_3)$. It is immediate on the fixed edges. For the remaining edges,because $f(e_{23})=[v_2{v}_1]=-e_{12}$. Finally,since $f(t)$ is degenerate and hence zero. Thus $h$ is the required chain homotopy.