Vector structure of $T_pM$ using the derivative of a chart

I think there is a flaw in this reasoning:

Let $M$ be an $m$-dimensional smooth manifold. We define the vector structure of $T_{p}M$ to be such that for a given (or really any - this does not depend on the chart chosen) chart $\phi$, its derivative $\phi^{*}(p)$ is linear. Since $\phi$ is a diffeomorphism, $\phi^{*}$ is known to be bijective (since it is a diffeomorphism), this suffices for it to be a isomorphism. Thus, with such a structure, $T_{p}M$ is isomorphic to $\mathbb{R}^{m}$.

The flaw being when I say that "Since $\phi$ is a diffeomorphism, $\phi^{*}$ is known to be bijective (since it is a diffeomorphism)". I would like to confirm that this is wrong, and know how to fix the argument, i.e how do we define the vector structure of $T_pM$ forcing the derivative of a chart to be linear.

• That statement is correct. – Qiaochu Yuan Mar 2 '17 at 23:03
• @QiaochuYuan Ok, so now my question is why is it true that $f$ diffeomorphic implies $f_*$ diffeomorphic. Actually this was the original question, but I found no reasons for this to be true – Soap Mar 2 '17 at 23:10
• A diffeomorphism is a smooth map with a smooth inverse. If $f$ is a diffeomorphism with inverse $g$, then the derivative of $f$ has inverse the derivative of $g$. – Qiaochu Yuan Mar 2 '17 at 23:13
• @QiaochuYuan I know that. I'm asking why are $f_*$ and its inverse smooth and bijective... – Soap Mar 2 '17 at 23:20
• @Simoes, I think you are a little confused. A diffeomorphism is a bijective map that os differentiable with inverse differentiable. – L.F. Cavenaghi Mar 2 '17 at 23:33

Let $M$ be an $m$-dimensional smooth manifold. We define the vector structure of $T_{p}M$

First, note that we are only defining a vector structure on $T_{p}M$, not a manifold structure (that too is possible, but the question is only about the argument presented here).

Since $\phi$ is a diffeomorphism, $\phi^{*}$ is known to be bijective (since it is a diffeomorphism)

The grammar is bad, but actually "it" refers to $\phi$, not $\phi^*$. And $\phi^*$ - or more accurately, $\phi_p^*$ since we are only talking about $T_pM$ - is bijective because that is part of the definition of $\phi$ being a diffeomorphism.

this suffices for it to be a isomorphism.

i.e., we define the vector structure on $T_pM$ to make $\phi_p^*$ a vector space isomorphism. (Note that it says "isomorphism", not "diffeomorphism".)

• Very well. Just one thing: you said that $\phi_*$ is bijective because that is part of the definition of $\phi$ being a diffeomorphism. But $\phi$ being a diffeomorphism only means that $\phi$ is bijective and both $\phi$ and its inverse are differentiable. It doesn't say anything about the bijectivity of $\phi_*:T_pM\rightarrow\mathbb{R}^m$ – Soap Mar 3 '17 at 14:48
• How directly you get it depends on exactly the definition of diffeomorphism used by whatever source you are studying (they are all equivalent, but different people prefer different versions), but it follows very easily from any version: since $\phi^{-1}\circ \phi = \text{id}$, by the chain rule $d\phi^{-1} \circ d\phi =d\text{id} = \text{id}$, so $d\phi$ has an inverse. – Paul Sinclair Mar 3 '17 at 18:12
• One thing: if $f:M\rightarrow N$ is a diffeomorphism and its derivative $f_*$ is bijective, then $f_*$, being linear, is an isomorphism. So we get an isomorphism between spaces with different dimensions... – Soap Mar 4 '17 at 19:44
• We would if such an $f$ existed. Therefore if $M$ is diffeomorphic to $N$, they must have the same dimension. – Paul Sinclair Mar 4 '17 at 19:54
• Then your "there are diffeomorphic functions $f: M \to N$" is simply FALSE when $M$ and $N$ are of different dimensions. There are no such functions, so it is hardly surprising that this absurd assumption gives you absurd results. – Paul Sinclair Mar 5 '17 at 23:20