I'm trying to prove the following.
If $f:M\to N$ is a diffeomorphism from an $m$-dimensional (non-empty) manifold to an $n$-dimensional manifold, then $m=n$.
My attempt: Fix a point $p\in M$ and consider the differential $(Df)_{p}:T_{p}M\to T_{q}N$, where $q=f(p)$. If we choose local coordinates $\{x^{1},\ldots,x^{m}\}$ centered at $p$, then $$\left\lbrace\frac{\partial}{\partial x^{1}}\bigg|_{p},\ldots,\frac{\partial}{\partial x^{m}}\bigg|_{p}\right\rbrace$$ is a basis for the tangent space $T_{p}M$. Similarly, if we pick local coordinates $\{y^{1},\ldots, y^{n}\}$ centered at $q=f(p)$, then $$\left\lbrace\frac{\partial}{\partial y^{1}}\bigg|_{q},\ldots,\frac{\partial}{\partial y^{n}}\bigg|_{q}\right\rbrace$$ is a basis for $T_{q}N$. In particular, we have $\dim(M)=\dim(T_{p}M)=m$ and $\dim(N)=\dim(T_{q}N)=n$. Since $f:M\to N$ is a diffeomorphism by assumption, the differential $(Df)_{p}:T_{p}M\to T_{q}N$ is an isomorphism. But isomorphic vector spaces must have the same dimension, so we may conclude that $m=n$.
My concerns: I think this okay, but it also seems too simple. Overall, does this proof work? Thanks in advance for any help!