# Topology - quotient spaces

Let X,Y be two spaces, A a closed subset of X, f:A--->Y a continuous map. We denote by $X\cup_fY$ the quotient space of the disjoint union $X\oplus{Y}$ by the equivalence relation ~ generated by $a ~ f(a)$ 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 $X\cup_fY$.

ii) The map f is extendable to a continuous map F:X-->Y if and only if Y is a retract of $X\cup_fY$.

I'm really stuck on this question. If A is a strong deformation retract of X, then we have some homotopy $H:X\times[0,1]\rightarrowX$ 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 $X\cup_fY$? :yuck:

Are the members of $X\cup_fY$ 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 $\varphi:X\cup_fY\rightarrow{Y}$ where for all y in Y, $\varphi(y) = y$.

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matt grime
Homework Helper
Things in the attachment are things in X or Y but where we identify an element in A with its image. Example: X=Y= a disc A is the boundary of X and f is an isomorphism onto the boudary of Y. The attachment is then two discs identified along their boundary, ie a sphere.

now imagine X is a punctured disc, so it is strongly homotopic to its boundary, S^1. Then it is clear that the join along the boundary is now a punctured sphere, and that this is homotopy equivalent to the disc Y, and that this homotopy can just be taken by extending the strong homotopy of X to its boundary by making it the identity on Y.

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hmm thanks a lot Matt I think I've got it.

matt grime