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**Additional info for Analysis and partial differential equations. Dedicated to Mischa Cotlar**

**Example text**

We first note that 0 E B because 0::; a for all a E A due to (i) so B satisfies (a). Let us now consider any non-empty linearly ordered subset C of B and let c = sup C. For an element a E A, we get either b ::; a for all b E B, in which case c ::5 a; or a ::5 f3 for some f3 E B, in which case a ::5 c. We thus fmd that c E B so that B satisfies (c) . To show that B also satisfies (b) we have to prove that f(b) E B for any given bE B, namely, we have to prove that either f(b) ::5 a or a ::5 f(b) for each a E A.

We thus obtain g = 1-1. 0 We now try to determine the necessary and sufficient conditions for the existence of either a left-inverse or a right-inverse of a function in case the function itself is not invertible. 3. A function 1 : X -+ Y has a left-inverse if and only if it is injective. For sufficiency let us assume that I is injective. In this case only a single element x E X corresponds to an element y = I(x) E n(f) ~ Y. Thus the function 1 : X -+ n(f) is bijective. 2 states that this function has an inverse I-I : n(f) -+ X.

I is a surjective function so that it has right-inverses. It is easily seen that I-l({a}) = {1,3}, rl({b}) = {2} . Therefore, we can defme two right-inverses: gl (a) = 1, gl (b) = 2 and g2(a) = 3, g2(b) = 2. One can check that both functions satisfy the relations I 0 gl = iy, I 0 g2 = iy. 2. 6 Inverse Functions 25 [~] ~ [~ -! ;] [::1 Since R(f) = Y = ]R2, the function I is surjective. Therefore, it has rightinverses. As a matter of fact, we can immediately see that the family of functions 9 : JR2 -+ JR3 represented by the following family of 3 x 2 matrices depending on two arbitrary real parameters a and b such that [1 [ :: k: R Xl 3 53a "8 - 16 corresponds indeed to an infmite family of right-inverses prescribed by the parameters a and b.