Solution

Assume (i), and suppose a map

f

as in (ii) existed. Then

A = f^{- 1} (I), B = f^{- 1} (J), 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) = \emptyset,

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) = dist (x, A), d_{B} (x) = dist (x, B), d_{C} (x) = 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

\frac{1}{D} (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).

Solved by gpt-5.6-sol high.

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