[proofplan] A Riemannian metric on $M$ is a smooth assignment $p \mapsto g_p$ of an inner product on each tangent space $T_pM$. We must verify three properties of $g := F^*h$: (i) for each $p \in M$, $g_p : T_pM \times T_pM \to \mathbb{R}$ is bilinear and symmetric; (ii) $g_p$ is positive-definite; and (iii) $g$ depends smoothly on $p$. Properties (i) and (iii) are inherited directly from $h$ — bilinearity, symmetry, and smoothness are preserved by pullback under any smooth map $F$. Positive-definiteness is the only place where the immersion hypothesis enters: it ensures $dF_p$ is injective, so that $h_{F(p)}(dF_p u, dF_p u) > 0$ for $u \neq 0$, since $h$ is positive-definite at $F(p)$ and $dF_p u \neq 0$. [/proofplan]

[step:Verify that $g_p$ is bilinear and symmetric for each $p \in M$] Fix $p \in M$ and let $u, v, w \in T_pM$, $\alpha, \beta \in \mathbb{R}$. By the linearity of $dF_p : T_pM \to T_{F(p)}N$ and the bilinearity of $h_{F(p)} : T_{F(p)}N \times T_{F(p)}N \to \mathbb{R}$, \begin{align*} g_p(\alpha u + \beta v,\, w) &= h_{F(p)}\bigl(dF_p(\alpha u + \beta v),\, dF_p w\bigr) \\ &= h_{F(p)}\bigl(\alpha\, dF_p u + \beta\, dF_p v,\, dF_p w\bigr) \\ &= \alpha\, h_{F(p)}(dF_p u,\, dF_p w) + \beta\, h_{F(p)}(dF_p v,\, dF_p w) \\ &= \alpha\, g_p(u, w) + \beta\, g_p(v, w). \end{align*} Linearity in the second argument is identical. For symmetry, since $h_{F(p)}$ is symmetric, \begin{align*} g_p(u, v) &= h_{F(p)}(dF_p u,\, dF_p v) = h_{F(p)}(dF_p v,\, dF_p u) = g_p(v, u). \end{align*} [/step]

[step:Verify that $g_p$ is positive-definite using the immersion hypothesis]Fix $p \in M$ and $u \in T_pM$ with $u \neq 0$. We must show $g_p(u, u) > 0$. By definition, \begin{align*} g_p(u, u) &= h_{F(p)}(dF_p u,\, dF_p u). \end{align*} Since $F$ is a smooth immersion, $dF_p : T_pM \to T_{F(p)}N$ is injective by [definition of immersion](/page/Immersion); in particular, $u \neq 0$ implies $dF_p u \neq 0$. Since $h_{F(p)}$ is an inner product on $T_{F(p)}N$, it is positive-definite, so $dF_p u \neq 0$ gives \begin{align*} h_{F(p)}(dF_p u,\, dF_p u) > 0. \end{align*} Hence $g_p(u, u) > 0$, and combined with the bilinearity and symmetry of Step 1, $g_p$ is an inner product on $T_pM$.[/step]

[guided]We must show that $g_p$ is positive-definite: for every nonzero $u \in T_pM$, the value $g_p(u, u)$ is strictly positive. This is the only place in the entire proof where the hypothesis that $F$ is an *immersion* — and not merely a smooth map — enters. Why is this hypothesis needed here, and not in Steps 1 or 3? Bilinearity, symmetry, and smoothness of the pullback $F^*h$ hold for any smooth $F$, with no rank condition on $dF$ required. Positive-definiteness, by contrast, is delicate: it can be destroyed by pullback whenever $dF_p$ has a kernel. So the immersion hypothesis is precisely the ingredient that prevents this collapse. Fix $p \in M$ and let $u \in T_pM$ with $u \neq 0$. Unfolding the definition of $g = F^*h$ at the pair $(u, u)$: \begin{align*} g_p(u, u) &= h_{F(p)}(dF_p u,\, dF_p u). \end{align*} We now ask: is the right-hand side positive? Since $h_{F(p)}$ is an inner product on $T_{F(p)}N$, it is positive-definite, so $h_{F(p)}(w, w) > 0$ for every nonzero $w \in T_{F(p)}N$. The strategy is therefore to show $dF_p u \neq 0$, then invoke positive-definiteness of $h$ at the point $F(p)$. Why is $dF_p u \neq 0$? This is exactly where the immersion hypothesis is consumed. By the [definition of immersion](/page/Immersion), $F$ being a smooth immersion means that $dF_p : T_pM \to T_{F(p)}N$ is injective at every $p \in M$. Injectivity of $dF_p$ means $\ker dF_p = \{0\}$, so the only vector in $T_pM$ that $dF_p$ sends to zero is $0$ itself. Since we assumed $u \neq 0$, injectivity gives $dF_p u \neq 0$. Setting $w = dF_p u$ in the positive-definiteness of $h_{F(p)}$: \begin{align*} h_{F(p)}(dF_p u,\, dF_p u) > 0. \end{align*} Combining the two displays, $g_p(u, u) > 0$. Together with the bilinearity and symmetry established in Step 1, this shows $g_p$ is an inner product on $T_pM$. It is worth seeing exactly what fails if we drop the immersion hypothesis. Suppose $F$ has a critical point at $p$ — that is, $dF_p$ is *not* injective. Then $\ker dF_p$ contains some nonzero $u \in T_pM$, and we compute \begin{align*} g_p(u, u) &= h_{F(p)}(dF_p u,\, dF_p u) = h_{F(p)}(0, 0) = 0. \end{align*} So $g_p(u, u) = 0$ for a nonzero $u$, meaning $g_p$ is only positive *semi*-definite, not positive-definite — it is a degenerate symmetric bilinear form, not an inner product. The immersion hypothesis (pointwise injectivity of $dF_p$) is precisely what converts positive-definiteness of $h$ at $F(p)$ into positive-definiteness of $F^*h$ at $p$. One subtle point: we did **not** need $F$ itself to be injective on $M$ — only the differential $dF_p$ at each point. This is strictly weaker (an immersion can fold $M$ onto itself, identifying distinct points of $M$ to a single image point in $N$, as long as no tangent vector is collapsed at any individual $p$). This is why the conclusion of the theorem is "$F^*h$ is a Riemannian metric on $M$" rather than "on $F(M)$": each tangent space $T_pM$ acquires its own genuine inner product through the local injectivity of $dF_p$, regardless of whether $F$ itself is globally one-to-one.[/guided]

[step:Verify the smooth dependence of $g$ on $p$] To show $g$ is smooth, we verify that for any smooth vector fields $X, Y \in \mathfrak{X}(M)$, the function $p \mapsto g_p(X_p, Y_p)$ is smooth. By the definition of $g$, \begin{align*} g_p(X_p, Y_p) &= h_{F(p)}(dF_p X_p,\, dF_p Y_p). \end{align*} Define the pushed-forward vector fields along $F$ by \begin{align*} \tilde X : M &\to TN, & p &\mapsto dF_p X_p, \\ \tilde Y : M &\to TN, & p &\mapsto dF_p Y_p. \end{align*} Both $\tilde X$ and $\tilde Y$ are smooth, since they are compositions of smooth maps: $X$ and $Y$ are smooth sections of $TM$, and $dF$ is a smooth bundle map $TM \to F^*TN$ (the differential of $F$). Moreover, $h$ is a smooth section of $T^*N \otimes T^*N$, so the function \begin{align*} M &\to \mathbb{R}, & p &\mapsto h_{F(p)}(\tilde X_p, \tilde Y_p) = (F^* h)(X, Y)_p \end{align*} is smooth as the composition of smooth maps. Hence $p \mapsto g_p(X_p, Y_p)$ is smooth for arbitrary smooth $X, Y$, which is the criterion for $g$ to be a smooth $(0,2)$-tensor field. [/step]

[step:Conclude] Combining the previous three steps: $g_p$ is bilinear, symmetric, and positive-definite at each $p \in M$, and $p \mapsto g_p$ is smooth. By [definition of Riemannian metric](/page/Riemannian%20Metric), $g = F^*h$ is a Riemannian metric on $M$. [/step]