Relative simplicial approximation after subdivision

Take $h$ and $T_r$ from the neighbourhood-adjustment lemma. Compose its homotopy $H_t$ with $f$ to obtain a homotopy from $f$ to $fh$, fixed on $\mathrm{\mid }A\mathrm{\mid }$. The inverse images under $fh$ of the target vertex stars form an open cover of compact metric $\mathrm{\mid }K\mathrm{\mid }$, so there is a Lebesgue number $\delta >0$. For empty $K$ the empty map already proves the theorem.

Choose $r\ge 3$ so that every star of a $T_r$ vertex in $\mathrm{\mid }{B}_2\mathrm{\mid }$ has diameter less than $\delta$. For each such vertex the Lebesgue-number property gives a target vertex $g(v)$ whose open star contains $fh(\mathrm{st}(v))$. For any remaining vertex $v$, the adjustment lemma gives $a\in A$ with $h(\mathrm{st}(v))\subseteq {\mathrm{st}}_A(a)$. Since $f{\mathrm{\mid }}_A$ is simplicial, a point having positive $a$ coordinate has positive $f(a)$ coordinate in its image, so $f({\mathrm{st}}_A(a))\subseteq {\mathrm{st}}_L(f(a))$; take $g(v)=f(a)$. On $A$ take $a=v$, hence $g(v)=f(v)$. These are finitely many choices.

All vertex stars now satisfy the criterion for $fh$, so $g$ extends simplicially and the criterion supplies $J_t(x)=(1-t)fh(x)+t\mathrm{\mid }g\mathrm{\mid }(x)$. On $\mathrm{\mid }A\mathrm{\mid }$, the simplicial maps $g$ and $f$ agree on vertices and therefore on all affine combinations; also $h$ is the identity there. Thus $J$ fixes $f{\mathrm{\mid }}_A$. Concatenate $f{H}_{2t}$ for $0\le t\le 1\mathrm{/}2$ with $J_{2t-1}$ for $1\mathrm{/}2\le t\le 1$. At the join both equal $fh$, so this is a continuous homotopy from $f$ to $\mathrm{\mid }g\mathrm{\mid }$, fixed on $A$. Its restriction there always lies in the simplicial image $f(A)$, proving the pair assertion.

To extend a prescribed compatible $A^{\mathrm{\prime }}$, process simplices of $K$ outside $A$ by increasing dimension. Their boundary triangulations are already fixed and compatible. Cone each such boundary from an interior barycenter of its original simplex. Every ray meets the boundary once, so these cones triangulate the simplex and restrict to the already specified boundaries. This finite induction constructs $K^{\mathrm{\prime }}$ restricting exactly to $A^{\mathrm{\prime }}$. If $f$ is simplicial there, apply the preceding argument to $(K^{\mathrm{\prime }},A^{\mathrm{\prime }})$. When $A$ is empty it is ordinary approximation; when $A=K$ and $f$ is already simplicial take $g=f,r=0$.