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

Question

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.

Answer 1

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".)