How to verify if a mapping is quotient.

[SVD]

Prove or disprove that f is a quotient mapping.
f:R^3\{(x1,x2,x3):x1=0}--->R^2 defined by

(x1,x2,x3)|->(x2/x1,x3/x1)

 

[genericusrnme]

Do you know what the definition of a quotient mapping is?
What properties should it have?

 

[Bacle2]

You usually need to have specific topologies defined in your domain and codomain, to test whether a map is a quotient map.

 

[Alesak]

I would try using directly the definition.

One direction is definition of continuity, what about the other direction?

I'm honestly not sure myself, but I would try to show that: at least one partial derivative in at least one coordinate is nonzero → it is localy injective → it is open → it is quotient.OT: I'm having trouble coming up with surjective continuous map between euclidean spaces that is not open. Is there some theorem about it? It would mean that every such map is quotient, it seems. If the map was not subjective, it would be easy.

Another OT, but vaguely related: If a countinuous map between euclidean spaces was open on any open subspace of domain, would it imply all that map would be open? There doesn't seem to be any way of continuously passing from open map on a subspace to a place where it takes open subset to non-open subset. So the OP would need to just check if some nice subspace is open.

 

[blue_raver22]

To verify if a mapping is a quotient, we need to check if it satisfies the definition of a quotient mapping. A quotient mapping is a surjective mapping from a topological space X to a quotient space Y such that the preimage of any open set in Y is open in X. In simpler terms, a quotient mapping is a continuous, onto mapping that preserves the topology of the original space.

To prove or disprove that f is a quotient mapping, we need to show that it satisfies the definition of a quotient mapping.

First, we need to check if f is a surjective mapping. This means that for every point in the target space (R^2), there exists a preimage in the domain space (R^3\{(x1,x2,x3):x1=0}). In this case, for any point (x,y) in R^2, we can find a preimage (x1,x2,x3) in R^3\{(x1,x2,x3):x1=0} by setting x1=1, x2=x and x3=y. Therefore, f is a surjective mapping.

Next, we need to check if f is continuous. To do this, we need to show that the preimage of any open set in R^2 is open in R^3\{(x1,x2,x3):x1=0}. Let U be an open set in R^2, then the preimage of U under f is given by {(x1,x2,x3)∈R^3\{(x1,x2,x3):x1=0}: (x2/x1,x3/x1)∈U}. Since U is open in R^2, we can find an open ball around each point in U. Let (a,b) be a point in U, then there exists an open ball B((a,b),r) contained in U. Now, consider the open ball B((a,b),(r/2)) in R^3\{(x1,x2,x3):x1=0}. This is the preimage of B((a,b),r) under f, which means that the preimage of any open set in R^2 is open in R^3\{(x1,x2,x3):x1=0}. Therefore, f is a continuous mapping.

Lastly, we need to show that f preserves the