# Three questions on injective functions.

1. Feb 27, 2009

### robbins

Q1. Claim: Suppose f : Rn -> Rm is injective. Then m >= n. Is this true?

Q2. Claim: Suppose f : Rn -> Rn is injective and f(X) = [f1(X) f2(X) ... fn(X)]T. Then each fk must be injective. Is this true?

Q3. I assume the above claims are known results or have known counterexamples. Can someone direct me to a good text or reference for questions such as these?

Thanks.

2. Feb 27, 2009

### CRGreathouse

3. Feb 27, 2009

### robbins

A space-filling curve is everywhere self-intersecting, and therefore can't be injective (though they are surjective). However, your point is taken: by Cantor's theorem the cardinality of [0, 1] is the same as [0, 1]^n for any finite n. And a mapping f : A -> B where |A| = |B| can be injective.

4. Feb 27, 2009

### yyat

All Rn's (n>0) have the same http://en.wikipedia.org/wiki/Cardinality" [Broken], hence there exist bijective maps between them.

Not sure if I understand you correctly. The identity map on R^2 is injective, but the components are the coordinate projections, which are not injective.

Last edited by a moderator: May 4, 2017
5. Feb 27, 2009

### CRGreathouse

Right, sorry. I knew that they were all the same size, and I knew about space-filling curves, but I didn't check that they actually worked for the problem stated!

6. Feb 27, 2009

### John Creighto

I think he means that if the infective mapping can be written in a form where the mapping for each component only depend on the value of that component, the mapping for each component is injective.

For such a statement the hypothesis is is difficult to satisfy because it almost looks like a one dimensional curve in an n dimensional space.

Last edited by a moderator: May 4, 2017
7. Feb 28, 2009

### robbins

Your interpretation is correct. My original question was not stated clearly (should have used subscripts).