By G. Lorentz

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**Example text**

1(i). 2 Let (W, F) be a scheme, and let 4' be a morphism from (X, G) to (W, F). Then we have (i) 14' = 1w. (ii) For each E 9 C(F), E4' -1 9 C. (iii) Assume that 4' is surjective. Then, for all d, e 9 F, (de)4' -~ = d4'- l e r -1. Proof. (i) Let z E X be given. Then (x, x) E 1. Therefore, (xr x4,) 9 1r On the other hand, as x4, 9 W, (z4,, xr 9 l w . It follows that 14, = l w . (ii) Let E 9 C(F) be given. Then 1w 9 E. Thus, by (i), 14' 9 E, which means that 1 9 E r In particular, E4,-1 ~= 0. Let c, d 9 E4, -1, and let g 9 c*d be given.

4(ii), gH E ( F ' / / H ) ( F / / H ) : (F//H)*(F//H) C_ F//H. 1 yields g E H F H C_ F. Since g E F*F has been chosen arbitrarily, we have shown that F E C. Conversely, let us now assume that F E C. Let g E G be such that gH E (F//H)*(F//H). 4(ii), we have (F//H)* = F*//H. Therefore, gH E (F*//H)(F//H). 1, g E (HF*H)(HFH) = (H*F*H*)(HFH) C_ F*F C F. It follows that gH E F//H. 24 1. Basic Results Since gH E (F//H)* (F//H) has been chosen arbitrarily, we have shown [] that F N H E C(G//H). 3 Let g E G, and let H E C be given.

Then F//U E C(G//H) if and only if F E C. 2 Proof. Assume first that F//H E C(G//H). Let g E F ' F be given. Then g E (HF*H)(HFH). 4(ii), gH E ( F ' / / H ) ( F / / H ) : (F//H)*(F//H) C_ F//H. 1 yields g E H F H C_ F. Since g E F*F has been chosen arbitrarily, we have shown that F E C. Conversely, let us now assume that F E C. Let g E G be such that gH E (F//H)*(F//H). 4(ii), we have (F//H)* = F*//H. Therefore, gH E (F*//H)(F//H). 1, g E (HF*H)(HFH) = (H*F*H*)(HFH) C_ F*F C F. It follows that gH E F//H.

### Approximation of Functions by G. Lorentz

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