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gazzo
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Let X,Y be two spaces, A a closed subset of X, f:A--->Y a continuous map. We denote by [itex]X\cup_fY[/itex] the quotient space of the disjoint union [itex]X\oplus{Y}[/itex] by the equivalence relation ~ generated by [itex]a ~ f(a)[/itex] for all a in A. This space is called teh attachment of X with Y along A via f.
i) If A is a strong deformation retract of X, show that Y is a deformation retract of [itex]X\cup_fY[/itex].
ii) The map f is extendable to a continuous map F:X-->Y if and only if Y is a retract of [itex]X\cup_fY[/itex].
I'm really stuck on this question. If A is a strong deformation retract of X, then we have some homotopy [itex]H:X\times[0,1]\rightarrowX[/itex] where for all x in X, H(x,0) = x. H(x,1) is in A. and for all b in A and all t in [0,1], H(a,t) = a.
But what has that got to do with [itex]X\cup_fY[/itex]?
Are the members of [itex]X\cup_fY[/itex] the equivalence classes where for a~b if one of: a = b, f(a)=f(b), a is in A and b=f(u) in Y.
So we want to come up with a map [itex]\varphi:X\cup_fY\rightarrow{Y}[/itex] where for all y in Y, [itex]\varphi(y) = y[/itex].
i) If A is a strong deformation retract of X, show that Y is a deformation retract of [itex]X\cup_fY[/itex].
ii) The map f is extendable to a continuous map F:X-->Y if and only if Y is a retract of [itex]X\cup_fY[/itex].
I'm really stuck on this question. If A is a strong deformation retract of X, then we have some homotopy [itex]H:X\times[0,1]\rightarrowX[/itex] where for all x in X, H(x,0) = x. H(x,1) is in A. and for all b in A and all t in [0,1], H(a,t) = a.
But what has that got to do with [itex]X\cup_fY[/itex]?
Are the members of [itex]X\cup_fY[/itex] the equivalence classes where for a~b if one of: a = b, f(a)=f(b), a is in A and b=f(u) in Y.
So we want to come up with a map [itex]\varphi:X\cup_fY\rightarrow{Y}[/itex] where for all y in Y, [itex]\varphi(y) = y[/itex].