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For contiguous simplicial maps, $|g|(x)$ and $|h|(x)$ lie in a common simplex of $L$ for every $x$. Its convexity makesa well-defined map into $|L|$, using the ambient Euclidean realization. The formula is continuous globally and agrees across simplex faces. At $t=0,1$ it equals $|g|,|h|$, so it is the desired homotopy.

A simplicial approximation $g:K\to L$ to $f:|K|\to |L|$ is a vertex map satisfying the open-star conditionfor every vertex $v$. Here an open star consists of points whose barycentric coordinate at that vertex is positive. This condition ensures that the images of the vertices of every simplex span a simplex of $L$, and the vertex map extends linearly over simplices.

To triangulate $|K|\times |L|$, order the vertices of each complex consistently. For each product $[v_0,\dots ,v_p]\times [w_0,\dots ,w_q]$, use the simplices whose vertices $(v_i,w_j)$ follow monotone lattice paths from $(0,0)$ to $(p,q)$, taking a step in one coordinate at a time. These prism triangulations agree on faces and cover the product.

If $g,h$ both approximate $f$, choose $x$ in the relative interior of a simplex $\sigma$. Then $x$ belongs to the open star of every vertex $v$ of $\sigma$. Hence $f(x)$ belongs to every open star of $g(v)$ and $h(v)$. Its unique carrier simplex in $L$ contains all these vertices. Thus $g(\sigma )$ and $h(\sigma )$ are faces of that one simplex, proving that $g,h$ are contiguous simplicial maps.

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.