The realization of a simplicial map is continuous and functorial

If $\sigma =\{v_0,\dots ,v_n\}$ is a simplex of $K$ and $x\in \mathrm{\mid }\sigma \mathrm{\mid }$ has barycentric coordinates $x={\sum }_{i=0}^n{\lambda }_i{v}_i$, then $\mathrm{\mid }f\mathrm{\mid }(x)={\sum }_{i=0}^n{\lambda }_{i}f(v_i)\mathrm{.}$ Thus the restriction $\mathrm{\mid }f\mathrm{\mid }{\mathrm{\mid }}_{\mathrm{\mid }\sigma \mathrm{\mid }}:\mathrm{\mid }\sigma \mathrm{\mid }\to \mathrm{\mid }f(\sigma )\mathrm{\mid }$ is the affine map determined by the vertex map $f$, so it is continuous.

For every barycentric function $\alpha$, the identity vertex map leaves every coefficient unchanged, so $\mathrm{\mid }{\mathrm{id}}_K\mathrm{\mid }(\alpha )=\alpha$. Likewise $\mathrm{\mid }g\circ f\mathrm{\mid }(\alpha )(u)={\sum }_{w\in g^{-1}(u)}{\sum }_{v\in f^{-1}(w)}\alpha (v)=\mathrm{\mid }g\mathrm{\mid }(\mathrm{\mid }f\mathrm{\mid }(\alpha ))(u)$, so realizations preserve composition.

If $\sigma$ and $\tau$ meet, then they meet along $\mathrm{\mid }\sigma \cap \tau \mathrm{\mid }$, and the affine formulas from step 1.1 agree there because they are both determined by the same vertex map $f$. Since $\mathrm{\mid }K\mathrm{\mid }$ carries the weak topology with respect to its simplices, these simplexwise affine maps patch to a continuous map $\mathrm{\mid }f\mathrm{\mid }:\mathrm{\mid }K\mathrm{\mid }\to \mathrm{\mid }L\mathrm{\mid }$.

Steps 2.1 and 1.2 give continuity and the identity/composition laws.