= Solution
Assume (i), and suppose a map $f$ as in (ii) existed. Then
$$
A=f^{-1}(I),\qquad B=f^{-1}(J),\qquad C=f^{-1}(K)
$$
are closed, cover $T$, and contain $I,J,K$, respectively. But
$$
A\cap B\cap C=f^{-1}(I\cap J\cap K)=\varnothing,
$$
since the three sides have no common point. This contradicts (i), so (i) implies (ii).
Conversely, suppose (i) fails. Let $A,B,C$ be a counterexample and put
$$
d_A(x)=\operatorname{dist}(x,A),\quad
d_B(x)=\operatorname{dist}(x,B),\quad
d_C(x)=\operatorname{dist}(x,C).
$$
No point lies in all three closed sets, so $D=d_A+d_B+d_C>0$. Identify $T$ affinely with the standard simplex so that side $I$ is the side on which the first barycentric coordinate is zero, and similarly for $J,K$. Define $f(x)$ to be the point with barycentric coordinates
$$
\frac1D(d_A(x),d_B(x),d_C(x)).
$$
Because $A\cup B\cup C$ covers $T$, at least one distance is zero, so $f(x)\in\partial T$. If $x\in I\subset A$, then $d_A(x)=0$, so $f(x)\in I$; likewise for $J,K$. Thus $f$ is a forbidden map from (ii). This proves (ii) implies (i).
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