Can a Riemannian metric always be induced by an immersion $F$? (I use $F$ as a local embedding)

Question

From this: Does the Riemannian metric induced by a diffeomorphism $F$ exist for a reason other than the existence of vector field pushforwards?

Proving a Riemannian metric has 2 parts: The inner product part and the smooth part.

• The answer there of Paulo Mourão seems to prove that the Riemannian metric can exist for $F$ only an immersion and does not use vector field pushforwards. It also seems that that $F$ is an immersion is used only for the inner product part, i.e. the smooth part is proven assuming only that $F$ is a smooth map.

• In this question Can a Riemannian metric always be induced by an immersion $F$? (I use that for any tangent vector, there exists a vector field), I try to prove that Riemannian metric can exist for $F$ immersion, and I think I also prove the smooth part assuming only that $F$ is a smooth map. I used Exercise 1.5.

Now, I attempt a different proof for the smooth part where $F$ is an immersion is used also in the smooth part. Question: Is this correct?

Let $X,Y \in \mathfrak X(N)$. We must show $\langle X,Y \rangle \in C^{\infty}N$. Smoothness is pointwise, so let us show $\langle X,Y$ is smooth at each $p \in N$. Let $p \in N$.

1. There exists a neighborhood $U_p$ of $p$ in $N$ such that $F|_{U_p}: U_p \to M$ is a smooth embedding, since immersions are equivalent to local embeddings.

2. (1) implies that $F(U)$ is a regular/an embedded submanifold of $M$ even if $F(N)$ is not open in $M$ (as would be the case for $F$ a local diffeomorphism) and even if $F(N)$ is not a regular/an embedded submanifold of $M$ (as would be the case for $F$ a local diffeomorphism onto image).

3. The pushforwards $F_{*}X, F_{*}Y$ are not necessarily defined since $F$ is not a diffeomorphism. Nevertheless, since $\tilde{F|_{U_p}}: U_p \to F(U_p)$ is a diffeomorphism, as shown in (2), we have for $G=\tilde{F|_{U_p}}$ that the pushforwards $G_{*,X}, G_{*,Y}$ are defined.

4. We can say that for $\langle X,Y \rangle'|_{U_p}: U_p \to \mathbb R$, we have that $\langle X,Y \rangle' = G^{*}\langle G_{*}X, G_{*}Y \rangle$ where $\langle G_{*}X, G_{*}Y \rangle$ is a map $\langle G_{*}X, G_{*}Y \rangle: G(U_p)=F(U_p) \to \mathbb R$ given by, for each $q \in U_p$ bijectively corresponding to each $G(q) \in G(U_p)$, $$(\langle G_{*}X, G_{*}Y \rangle)(G(q)) = \langle (G_{*}X)(G(q)), (G_{*}Y)(G(q)) \rangle_{G(q)} = \langle (G_{*}X)_{G(q)}, (G_{*}Y)_{G(q)} \rangle_{G(q)} = \langle G_{*,q} X_q, G_{*,q} Y_q \rangle_{G(q)}$$

5. $\langle X,Y \rangle'|_{U_p}$ is smooth at $p$ by the composition of smooth maps given in (4).

6. The restriction $\langle X,Y \rangle'|_{U_p}$ is smooth at $p$ if and only if the original $\langle X,Y \rangle'$ is smooth at $p$.

7. Therefore, by (5) and (6), $\langle X,Y \rangle'$ is smooth at $p$.

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Context: The motivation for doing this kind of proof is based on what I believe was the intended idea for the comments of user10354138 and of lEm here.

Answer 1

I think your proof is not complete. What your argument shows is that you can reduce to the case that $N$ is an embedded submanifold of $M$ and $F$ is the inclusion.

The problem is that you used that

$$\langle G_{*}X, G_{*}Y \rangle: G(U_p) \to \mathbb R$$ is smooth. Here $G_{*}X, G_{*}Y$ are smooth vectorfields on $G(U_p)=F(U_p)$. So for the above to make sense you have to restrict the metric $\langle \cdot,\cdot \rangle$ on $M$ to the submanifold $F(U_p)$. Then for your argument to work you have to make sure that the restricted metric is still smooth. This is the same as saying that the inclusion $i:F(U_p)\to M$ induces a smooth metric on $F(U_p)$.