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

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 $$p$$.

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.

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.
$$\langle G_{*}X, G_{*}Y \rangle: G(U_p) \to \mathbb R$$ 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 $$F(U_p)$$.