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Therefore T x = T y for all x, y ∈ X, i.e., T is a constant map.
Let be a map with zero degree and let be the constant map at.
So we suppose such that Such a choice is permissible as is not a constant map.
Finally, H 0 ( x ) is a constant map mapping each function x ∈ U ¯ to ℓ ( t ).
If we consider a constant map in Theorem 2.1, then we obtain the following result.
Now, by the proof of Theorem 4.1 of [9], for each, either or is homotopic to a constant map.
Similar(36)
Now define a constant mapping by, then for (3.35).
Thus, if is the constant mapping for all, the is not required to be locally convex.
If f ≡ u, the constant mapping on K, then Corollary 3.2 reduces to the following corollary.
If (B:overline{U}to U) is a constant mapping, then (i(B,U,V =1).
Therefore, N and M are all homeomorphic to the graph of a constant mapping.
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