Define a <simplicial map> $j:K\to L$ on vertices by
$$
j(v_1)=v_1,qquad j(v_2)=v_2,qquad
j(v_3)=v_1,qquad j(v_4)=v_4.
$$
The image of every simplex is a simplex of $L$, and $j\circ i=\operatorname{id}_L$. The maps $i\circ j$ and $\operatorname{id}_K$ 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). Define
$$
h_0(v_3)=e_{13},
\qquad h_0(v_1)=h_0(v_2)=h_0(v_4)=0,
$$
$$
h_1(e_{23})=t,
\qquad h_1(e_{12})=h_1(e_{13})=h_1(e_{14})=h_1(e_{24})=0,
\qquad h_2=0.
$$
We verify
$$
\partial h+h\partial=\operatorname{id}-f.
$$
It 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,
$$
(\partial h+h\partial)e_{13}=h_0(v_3-v_1)=e_{13}
=e_{13}-f(e_{13}),
$$
$$
(\partial h+h\partial)e_{23}
=\partial t+h_0(v_3-v_2)
=(e_{23}-e_{13}+e_{12})+e_{13}
=e_{23}+e_{12}
=e_{23}-f(e_{23}),
$$
because $f(e_{23})=[v_2v_1]=-e_{12}$. Finally,
$$
(\partial h+h\partial)t=h_1(e_{23}-e_{13}+e_{12})=t=t-f(t),
$$
since $f(t)$ is degenerate and hence zero. Thus $h$ is the required chain homotopy.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 20F / c / Solution

Simplicial maps $f,g:K\to L$ are contiguous when $f(\sigma)\cup g(\sigma)$ lies in one simplex of $L$ for every simplex $\sigma$ of $K$. Contiguous maps have homotopic geometric realizations.


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 Incoming links: Contiguous simplicial maps