By Joseph J. Rotman

**Publish yr note:** First released in 1988

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A transparent exposition, with routines, of the fundamental rules of algebraic topology. compatible for a two-semester direction at the start graduate point, it assumes a data of aspect set topology and uncomplicated algebra. even though different types and functors are brought early within the textual content, over the top generality is kept away from, and the writer explains the geometric or analytic origins of summary strategies as they're brought.

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**Example text**

PROOF. A minor variation of the proof just given. D Definition. An ordered set of points {Po, Pl' ... , Pm} C Rn is affine independent if {Pl - Po, P2 - Po, ... , Pm - Po} is a linearly independent subset of the real vector space Rn. Any linearly independent subset of R n is an affine independent set; the converse is not true, because any linearly independent set together with the origin is affine independent. Anyone point set {Po} is affine independent (there 33 Affine Spaces are no points of the form Pi - Po with i #- 0, and 0 is linearly independent); a set {Po, Pl} is affine independent if Pl - Po #- 0, that is, if Pl #- Po; a set {Po, Pl' P2} is affine independent if it is not collinear; a set {Po, Pl' P2' P3} is affine independent if it is not coplanar.

On each triangle (2-simplex), H shall be an affine map; it thus suffices to evaluate H on each vertex (observe that agreement on overlaps is automatic here). Define H(a) = H(q) = 0(, H(b) = H(p) = f3; H(c) = y; H(d) = b; H(r) = p. 8, the vertical edge [a, q] collapses to 0(, and the vertical edge [b, p] collapses to f3. Also, [q, d] goes to [0(, b], Ed, c] goes to [b, y], and [c, p] goes to [y, f3]. *(IPI of))d- 1 rell Therefore IPo*[f] = [IPo of] = [A * IPl of * rl] (using homotopy associativity).

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. Then f is an identification if and only if, for all spaces Z and all functions g: Y - Z, one has g continuous if and only if gf is continuous. L X Z. ~;. 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.