An implicit function theorem: Comment

In Ref. 1, Jittorntrum proposed an implicit function theorem for a continuous mappingF:R n ×R m →R n, withF(x 0,y 0)=0, that requires neither differentiability ofF nor nonsingularity of ∂ x F(x 0,y 0). In the proof, the local one-to-one condition forF(·,y):A ⊂R n →R n for ally ∈B is consciously or unconsciously treated as implying thatF(·,y) mapsA one-to-one ontoF(A, y) for ally ∈B, and the proof is not perfect. A proof can be given directly, and the theorem is shown to be the strongest, in the sense that the condition is truly if and only if.