My book is An Introduction to Manifolds by Loring W. Tu.

I think that since diffeomorphisms are equivalent to surjective smooth embeddings, I think this is the same as proving an injective local diffeomorphism is a smooth embedding (I understand "image is submanifold and diffeomorphism onto its image" is an equivalent definition of a smooth embedding) which is asked here and here (and here in the continuous case), but I would like a verification of my own proof:

  1. Let $M$ and $N$ be smooth manifolds of respective dimensions $m$ and $n$. Let $F: N \to M=F(N)$ be a bijective smooth map. Suppose $F$ is a local diffeomorphism. Let us show that $F^{-1}: M \to N$ is smooth, using Definition 6.5, to show that $F$ is actually a diffeomorphism.

  2. Let $U$ be open in $N$. I'll denote by

    • $F|_U$ as the domain-restriction $F|_U:U \to M$,

    • $\tilde{F|_U}$ as the range-restriction $\tilde{F|_U}:U \to F(U)$

    • $G$ as the inverse of $\tilde{F|_U}$, i.e. $G := (\tilde{F|_U})^{-1}: F(U) \to U$.

      • Equivalently, $G = \tilde{F^{-1}|_{F(U)}}: F(U) \to U$, the range-restriction of the domain-restriction $F^{-1}|_{F(U)}: F(U) \to N$ of $F^{-1}: M \to N$
  3. I understand that the "$\varphi \circ F^{-1} \circ \psi^{-1}$" in Definition 6.5 refers to "$\varphi \circ G \circ \psi^{-1}$"

  4. For all $F(p) \in M$, with $p \in N$, we must find a chart $(U, \varphi)$ about $p$ in $N$ and a chart $(V, \psi)$ about $F(p)$ in $M$ such that $$\varphi \circ G \circ \psi^{-1}: \psi(F(U) \cap V) \to (\varphi \circ G)(F(U) \cap V)$$ is smooth about the point $\psi(F(p))$ (its value at $\psi(F(p))$ is $\varphi(p)$).

  5. Let us now use the local diffeomorphism property: For all $p \in N$, there exists a neighborhood $U_p$ of $p$ in $N$ such that for the domain-restriction $F|_{U_p}: U_p \to M$ and the range-restriction $\tilde{F|_{U_p}}: U_p \to F(U_p)$, $F(U_p)$ is open in $M$ and $\tilde{F|_{U_p}}$ is a diffeomorphism.

  6. I'm not sure if $U_p$ has a coordinate map $\gamma$ that makes $(U_p,\gamma)$ into a chart, but since $U_p$ is an open subset of $N$, $U_p$ can be made into a smooth manifold. Viewing $p \in U_p$, there exists a chart $(A_p,\varphi_A)$ about $p$ in $U_p$. I know $A_p$ is open in both $U_p$ and $N$, and I think $(A_p,\varphi_A)$ is also a chart about $p$ in $N$. (If relevant: I guess we use that open subsets are equivalent to regular submanifolds of codimension zero.)

  7. $n=m$ because

    • $n = \dim N = \dim U_p = \dim F(U_p) = \dim M = m$, where

      • $\dim F(U_p)$ is defined because $F(U_p)$ becomes a manifold because $F(U_p)$ is an open subset of $M$ (from (5))

      • $\dim U_p = \dim F(U_p)$ because of (3) in another post

  8. I'll choose $(U, \varphi) = (A_p,\varphi_A)$ and $(V, \psi) = (G^{-1}(A_p), \varphi_A \circ G)$. This works because

    • 8.1 $$\varphi \circ F^{-1} \circ \psi^{-1} = \varphi \circ G \circ \psi^{-1} = \varphi_A \circ G \circ (\varphi_A \circ G)^{-1}$$

      $$ = \varphi_A \circ G \circ G^{-1} \circ \varphi_A^{-1} = \varphi_A \circ \varphi_A^{-1}$$

    • 8.2 $\psi(F(U) \cap V) = (\varphi_A \circ G)(F(A_p) \cap G^{-1}(A_p)) = (\varphi \circ G)(F(U) \cap V)$ (which turns out to be equal to $\varphi_A(A_p)$)

    • 8.3 By (8.2) and (7), it makes sense to say that (8.1) shows that $\varphi \circ F^{-1} \circ \psi^{-1}$ is an identity map on an open subset of $\mathbb R^m$.

  9. Identity maps on open subsets of $\mathbb R^m$ are smooth.

  10. Therefore, by (8) and (9), $F^{-1}$ is smooth.

  11. Therefore, by (10) and (1), $F$ is a diffeomorphism.

  • 2
    $\begingroup$ Try using the Proof Verification tag, you might get more attention $\endgroup$ Jul 20, 2019 at 12:28
  • $\begingroup$ @theREALyumdub THANKS! $\endgroup$
    – user636532
    Jul 20, 2019 at 12:29

1 Answer 1


Your proof is fine. But unless I am missing something, or using a different definition of "locally smooth", I would simply argue as follows: $F:M\to N$ is bijective, so it has an inverse $F^{-1}:N\to M$. Since $F$ is a local diffeomorphism, for each $p\in M,\ F^{-1}$ is differentiable at $F(p).$ It follows that $F$ is a diffeomorphism.

  • $\begingroup$ Thanks. I assume you mean by "differentiable" you mean "smooth". How do you prove that "Since $F$ is a local diffeomorphism, for each $p\in M,\ F^{-1}$ is differentiable (smooth) at $F(p)$" ? I think I ended up proving something like that here and in doing so relied on this very question. Something's up. It seems that something that would be circular actually isn't (well, I hope so). $\endgroup$
    – user636532
    Jul 21, 2019 at 6:12
  • 1
    $\begingroup$ Yes, differentiable= smooth, i.e. $\psi\circ F\circ \phi^{-1}$ is differentiable for appropriately chosen charts. $F^{-1}$ is smooth $by definition$ of diffeomorphism. The point is, differentiability is a local property to begin with, so the result is more or less automatic, once you have that additionally, $F$ is bijective. $\endgroup$ Jul 21, 2019 at 17:31
  • $\begingroup$ Thanks. Actually 1. Is this true? "A map $G: P \to Q$ of smooth manifolds is smooth if each $r \in P$ has a neighborhood $V_r$ in $P$ such that $G: V_r \to Q$ is smooth." I seem to have written it in another post but am not sure where I got that. This sounds like what you meant by "differentiability is a local property". (That differentiability is local is a concept is somewhere in Section 19 I think, but I can't find exactly anything like the quoted sentence) $\endgroup$
    – user636532
    Jul 23, 2019 at 3:55
  • $\begingroup$ 2. Can we also prove like this? Bijective immersions are equivalent to diffeomorphisms, and bijective local diffeomorphisms are equivalent to bijective immersions because either condition implies equal dimensions. This proof, if correct, sounds like what you're doing. (I hope this is acceptable as a comment because I am trying to understand this answer instead of ask a completely different question) $\endgroup$
    – user636532
    Jul 23, 2019 at 3:55

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