By Iain Adamson

This e-book has been known as a Workbook to make it transparent from the beginning that it's not a standard textbook. traditional textbooks continue through giving in each one part or bankruptcy first the definitions of the phrases for use, the innovations they're to paintings with, then a few theorems regarding those phrases (complete with proofs) and at last a few examples and routines to check the readers' knowing of the definitions and the theorems. Readers of this e-book will certainly locate the entire traditional constituents--definitions, theorems, proofs, examples and exercises yet now not within the traditional association. within the first a part of the publication could be came upon a short overview of the elemental definitions of basic topology interspersed with a wide num ber of workouts, a few of that are additionally defined as theorems. (The use of the be aware Theorem isn't really meant as a sign of hassle yet of significance and value. ) The routines are intentionally no longer "graded"-after the entire difficulties we meet in mathematical "real life" don't are available in order of trouble; a few of them are extremely simple illustrative examples; others are within the nature of educational difficulties for a conven tional path, whereas others are particularly tough effects. No strategies of the routines, no proofs of the theorems are integrated within the first a part of the book-this is a Workbook and readers are invited to aim their hand at fixing the issues and proving the theorems for themselves.

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**Additional resources for A General Topology Workbook**

**Sample text**

Y PROOF. Assume f is an identification. If g is continuous, then gf is continuous. Conversely, let gf be continuous and let V be an open set in Z. Then (gf)-l(V) = f-1(g-1(V)) is open in X; since f is an identification, g-l(V) is open in Y, hence g is continuous. Assume the condition. 3. Note that cp is surjective because f is. Consider the commutative diagram X ~ X/kerf ~ ;'-1 Y That cp-1f = v is continuous implies that cp-1 is continuous, by hypothesis. Also, cp is continuous because v is an identification.

A continuous surjection f: X - Y is an identification if a subset U of Y is open if and only if f-1(U) is open in X. 4. If '" is an equivalence relation on X and X/ '" is given the quotient topology, then the natural map v: X - X / '" is an identification. 5. If f: X - Y is a continuous surjection that is either open or closed, then f is an identification. 6. , there is a continuous s: Y - X with fs = 1y), then f is an identification (note that f must be a surjection). 8. Let f: X - Y be a continuous surjection.

Let A c Rn be an affine set and let T: A -+ Rk be an affine map. If X c A is affine (or convex), then T(X) c Rk is affine (or convex). In particular, if a, bare distinct points in A and if t is the line segment with endpoints a, b, then T(t) is the line segment with endpoints T(a), T(b) if T(a) #- T(b), and T(t) coJlapses to the point T(a) if T(a) = T(b). 9. If {Po, Pl' ... , Pm} is affine independent with barycenter b, then {b, Po, ... , Pi' ... , delete Pi) is affine independent for each i. 10.