- #1
Joey_Joe_Jojo
- 7
- 0
Hi, just a quick question concerning the invertibility of multivalued functions. Specifically, I am looking at
y=x^2
as a simple example. So for an inverse to exist we have to restrict the domain to (0,+infty) right (doesnt matter which branch I am taking, so Ill take this one.)
Now my problem is the point 0, because the point (0) is actually a multiset. For
y=x^2
there is a double zero. So formally speaking the mapping of the origin is (0,0) -> (0). Is this correct? Because if it is, then this implies that the point y=0 has no inverse, regardless of our choice of branch, because there is no bijection at this point.
That is my query. The point (0) is a branch point isn't it? So I am thinking that the inverse doesn't exist here. Is this correct? I've looked online and many sources say that if you restrict the domain to [0,+infty) you get an acceptable inverse, but I think it should be (0,+infty).
Any help that anyone could give would be excellent. I am not being pedantic, this is important for what I am doing. Also sorry if some of my maths definitions are ambiguous; I am a physicist not a mathematician!
Thank you,
Stephen
y=x^2
as a simple example. So for an inverse to exist we have to restrict the domain to (0,+infty) right (doesnt matter which branch I am taking, so Ill take this one.)
Now my problem is the point 0, because the point (0) is actually a multiset. For
y=x^2
there is a double zero. So formally speaking the mapping of the origin is (0,0) -> (0). Is this correct? Because if it is, then this implies that the point y=0 has no inverse, regardless of our choice of branch, because there is no bijection at this point.
That is my query. The point (0) is a branch point isn't it? So I am thinking that the inverse doesn't exist here. Is this correct? I've looked online and many sources say that if you restrict the domain to [0,+infty) you get an acceptable inverse, but I think it should be (0,+infty).
Any help that anyone could give would be excellent. I am not being pedantic, this is important for what I am doing. Also sorry if some of my maths definitions are ambiguous; I am a physicist not a mathematician!
Thank you,
Stephen