4

I read from Milnor's book $\textit{Topology from the Differentiable Viewpoint}$ this assertion

"If $f$ is a diffeomorphism between opensets $U\subset R^k$ and $V\subset R^l$, then k must equal l, and the linear mapping

$$df_x:R^k\rightarrow R^l$$ must be nonsingular."

The proof was: The composition $f^{-1}\circ f$ is the identity map of U; hence $d(f^{-1})_v\circ df_x$ is the identity map of $R^k$. Similarly $df_x \circ d(f^{-1})_v$ is the identity map of $R^l$. Thus $df_x$ has a two-sided inverse, and it follows that $k=l$.

My question is as follows:

I don't understand why the fact that $df_x$ has a two-sided inverse implies $k=1$. The way I would prove it is as follows: Instead of proving $df_x$ has a two-sided inverse, I would say one-sided inverse will suffice. Since $d(f^{-1})_v\circ df_x$, it follows that $df_x$ must be an isomorphism(Structure preserving comes from linearity. And bijectivity comes from the identity map.). Therefore $k=l$.

Thanks so much for your help!

Evariste
  • 1,617
  • I honestly don't understand why you don't want to use the fact that $df_x$ has a two sided inverse: it follows immediatly from the fact that $f$ and $f^{-1}$ are inverse functions... I think you are drowning in an empty cup of tea. – Mariano Suárez-Álvarez May 23 '13 at 06:49
  • 1
    Hi Mariano(or should I call you Prof. Suarez-Alvarez), thanks for the comment. I admit that not using the method you(and Brian) suggested sounds silly. I was just being pedantic and trying to figure out if I can get away with only one-sided inverse. Thanks for the help! – Evariste May 23 '13 at 06:57
  • 1
    Please call me Mariano! – Mariano Suárez-Álvarez May 23 '13 at 07:21

2 Answers2

4

It is not true that if a linear map $f:V\to W$ has a one sided inverse, then $\dim V=\dim W$ nor that $f$ is an isomorphism.

Mariano Suárez-Álvarez
  • 139,300
  • Here is my proof(please feel free to point out if I'm wrong):

    $\cdot$ $\phi(x+y)=\phi(x)+\phi(y)$ by linearity

     – Evariste May 23 '13 at 05:59
  • You must be wrong, independently of what your proof is :-) Consider the map $f:(x,y)\in\mathbb R^2\mapsto x\in\mathbb R$; it is easy to see that it has a one-sided inverse. On the other hand, the map $g:x\in\mathbb R\mapsto(x,0)\in\mathbb R^2$ has a one-sided inverse on the other side. Neither of these maps has comain and codomain of the same dimension. – Mariano Suárez-Álvarez May 23 '13 at 05:59
  • Here is my proof(please feel free to point out if I'm wrong):

    $\cdot$ $\phi(x+y)=\phi(x)+\phi(y)$ by linearity

    $\cdot$ $\phi(nx)=n \phi(x)$ by linearity

    $\cdot$ $d\phi$ being bijective: suppose that $d\phi$ is not bijective, then there exists elements $x, y \in V$ such that $d\phi(x)=d\phi(y)=a$. Then since $d\phi^{-1}\circ d\phi=id$, we have $d\phi^{-1}(a)=x$ and $d\phi^{-1}(a)=y$ at the same time, which contradict with the fact that $d\phi^{-1}$ is a linear map.

     – Evariste May 23 '13 at 06:05
  • Look at the two examples I gave in a comment above. (If you know that $V$ and $W$ are vector spaces of the same dimension, then it is true that a linear map $f:V\to W$ which has a one-sided inverse —on whatever side— is necessarily an isomorphism; but this does not apply to your case precisely because you are trying to prove that the two spaces have the same dimension) – Mariano Suárez-Álvarez May 23 '13 at 06:05
  • Thanks Mariano! However I don't think either example you've given can serve as a counter example. for the $x \rightarrow (x,0)$ example you've given, in order to define a one-sided inverse, we have to require the domain of $f^{-1}$ to be the entire $R^2$. However $R^2$ is larger than the range of $f$. Therefore we can't construct such a one-sided inverse. – Evariste May 23 '13 at 06:30
  • That example admits a one-sided inverse *on the other side, as I wrote above. – Mariano Suárez-Álvarez May 23 '13 at 06:34
  • Hi Mariano, sorry I think you were right. I was just being silly... Your counter example was right. And there was a loophole in my argument. i.e. I didn't really prove bijectivity, instead I only proved injectivity... Thanks a lot for your help and patience! – Evariste May 23 '13 at 07:25
2

It is a standard fact from elementary set theory that a map $f$ is a bijection $\iff$ $f$ has a two-sided inverse. Remember, $df_x$ is a linear map, and therefore, an isomorphism. This shows $k = l$ since isomorphic vector spaces have equal dimension.

Ink
  • 5,234
  • Thanks Brian. It's a good answer. However my main problem to be addressed was that if we really need a two-sided inverse? Please see Mariano's comment(if you're interested or want to help). I appreciate your effort. – Evariste May 23 '13 at 06:33