# How does P:R→{a,b,c}P:\Bbb R\rightarrow \{a,b,c\}induce a quotient topology

Let pp be a map of the real line onto the three point set A={a,b,c}A=\{a,b,c\} .

The map P:X→{a,b,c}P:X \rightarrow \{a,b,c\} defined by

p(x)={a,if x>0b,if x<0c,if x=0p(x)=\begin{cases} a,&\text{if }x>0\\
b,&\text{if }x<0\\ c,&\text{if }x=0 \end{cases} induces a quotient topology on AA. How does this induce a quotient topology? I believe I am supposed to first prove that the map pp is a quotient map but I don't know how to go about it. Could you please help out. ================= ================= 1 Answer 1 ================= By definition the quotient topology on AA induced by pp is the topology τ\tau such that for any U⊆AU\subseteq A, U∈τU\in\tau if and only if p−1[U]p^{-1}[U] is open in R\Bbb R. There are only 88 subsets of AA, so you can easily find their inverse images under pp and see which have open inverse images. Then just check whether they form a topology on AA. In other words, you donâ€™t start by proving that pp is a quotient map. If pp is to be a quotient map, we know exactly which subsets of AA should be open, and we just have to check that they really do form a topology on AA. While youâ€™re checking, try to see why you will always get a topology in this way.      You don't really need to check that it's a topology, do you? Any τ\tau formed in such a way is always a topology. – arkeet 2 days ago      @arkeet: Provided that the fibres of pp partition the domain, yes, and that is the case here. However, the OP is probably just starting to learn about quotient maps, so I think that checking that we get a topology would be instructive. – Brian M. Scott 2 days ago      How would they not partition the domain? Preimages preserve intersections. I agree that it would be instructive to list the elements of τ\tau and check that it is indeed a topology. – arkeet 2 days ago      @arkeet: If it turned out that pp was defined on a proper subset of the original space (here R\Bbb R). – Brian M. Scott 2 days ago 1   @arkeet: So do I, certainly in this context. There are some contexts in which partial functions might arise, and Iâ€™ve seen this notation abused in that way on very rare occasions, but thatâ€™s really pretty much out of the question here. I wasnâ€™t thinking so much of that possibility as I was of the possibility that the definition of the function was defective. That I definitely have seen! – Brian M. Scott 2 days ago