Rank Theorems and Embedded Submanifolds Examples

The coordinate inclusion $j$ is linear, so directly from the definition of the total derivative its differential at every point is $j$ itself. Its Jacobian is the injective block matrix $\left[\begin{array}{c}{I}_m\\ 0\end{array}\right]$, in agreement with [L1]. Hence $j$ is an immersion by [F1].

The projection $\pi$ is linear, so its differential at every point is $\pi$ itself. Its Jacobian is the surjective block matrix $\left[\begin{array}{cc}{I}_n& 0\end{array}\right]$, in agreement with [L1]. Hence $\pi$ is a submersion by [F1].

This verifies the example.

Let $A\in SL(n,\mathbb{R})$ and $H\in M_n(\mathbb{R})$. Using [F1], for small $t$ one has $\det (A+tH)=\det (A)\det (I+t{A}^{-1}H)=\det (I+t{A}^{-1}H)$ because $\det (A)=1$. Differentiating at $t=0$ with [L2] gives $D(\det {)}_A(H)=\mathrm{tr}(A^{-1}H)$.

This linear functional is surjective, because $D(\det {)}_A(A\mathrm{/}n)=\mathrm{tr}(I\mathrm{/}n)=1$. Hence $1$ is a regular value of the determinant.

The level set is nonempty because $\det I_n=1$. By [L1], ${\det }^{-1}(1)=SL(n,\mathbb{R})$ is therefore an embedded codimension-one submanifold.

The map $\pi$ is a submersion because its Jacobian is $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]$. For $g(u,v)=u^2+v^2$, [L3] gives $D{g}_{(u,v)}(a,b)=2ua+2vb$, which is surjective whenever $u^2+v^2=1$. The fibre is nonempty because $(1,0)\in g^{-1}(1)$. Hence $S^1=g^{-1}(1)$ is an embedded submanifold by [L2].

Applying [L1] to the submersion $\pi$ and the embedded circle $S^1$ gives that ${\pi }^{-1}(S^1)$ is an embedded submanifold of ${\mathbb{R}}^3$.

The displayed equation identifies this preimage with the usual cylinder.

Repeatedly applying the derivative identities in [L2] shows that every derivative of $f(x)=\sin x$ exists and is again $\pm \sin x$ or $\pm \cos x$; these functions are continuous. Hence $f$ is smooth.

Therefore its graph is an embedded submanifold of $\mathbb{R}\times \mathbb{R}\cong {\mathbb{R}}^2$ by [L1].

This is exactly the displayed subset $\mathrm{\Gamma }$.

The image of $f$ is $(0,\mathrm{\infty })$, so $f^{-1}(0)=\mathrm{\varnothing }$.

By [F1], regularity of $0$ asks whether every point of the empty fibre is regular, which is vacuously true. The derivative fact [L1] is compatible with that conclusion because the map has no critical point over $0$.

Hence $0$ is a regular value although it is not attained.

By [L1], ${\gamma }^{\mathrm{\prime }}(t)=(\cos t,2\cos 2t)$, and as in the false-statement proof this never vanishes. Hence $\gamma$ is an immersion.

The image passes through $(0,0)$ at $t=0$ and $t=\pi$, and in fact at every integer multiple of $\pi$. The two branches coming from $t=0$ and $t=\pi$ already have distinct tangent directions, so the image has a transverse self-crossing and is not locally homeomorphic to an interval at the origin.

Thus $\gamma (\mathbb{R})$ is an immersed image that is not an embedded submanifold, refuting [F1].

By [L1], $M$ is a smooth manifold. The restriction of $F$ to each component is the ordinary circle inclusion, so $F$ is an injective immersion.

In the intrinsic topology of $M$, each component is open. In the subspace topology on $F(M)$, the unit circle component is not open because every neighbourhood of one of its points meets infinitely many outer circles. So $F:M\to F(M)$ is not a homeomorphism and the topologies do not agree.

Therefore this single example simultaneously refutes both claims in [F1].

By [L1], the derivative at $0$ is $0$, so the rank at $0$ is $0$.

For every $x\mathrm{\ne }0$, the derivative is $2x\mathrm{\ne }0$, so the rank is $1$ at $x$. Thus no neighbourhood of $0$ has constant rank $0$.

Hence $f$ refutes the claim in [F1].