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The idea of Anton Petrunin can be made into an accurate proof. One does not need $C^2$ convergence, $C^1$ convergence is enough. That is, I claim that there is no $C^1$ immersion sufficiently $C^1$-close to the composition $\phi:S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$. (By the way, any map $S^2\to\Bbb R^3$ is $C^0$-close to a $C^\infty$ immersion, according to the $C^0$-dense $h$-principle and using that $S^2$ immerses in $\Bbb R^3$.)

Let $f:S^2\to\Bbb R^3$ be a self-transverse map (not necessarily an immersion) that is $C^1$-close to $\phi$. The image of $f$ lies in a tubular neighborhood $S^2\times\Bbb R$ of the image of $\phi$. Consider the composition $\psi:S^2\xrightarrow{f}S^2\times\Bbb R\xrightarrow{\text{projection}}\Bbb R$R\xrightarrow{\text{projection}}S^2$. It is$C^1$-close to$\phi$, so it is equivalent to$\phi$by a change of coordinates outside a small neighborhood of the poles (which are the singular points of$\phi$). So we may assume that, outside of a small neighborhood of the poles,$f$is a vertical lift of$\phi$(with respect to the projection$S^2\times\Bbb R\to S^2$). Then, in particular,$f$sends the equator of$S^2$into the plane$\Pi$in$\Bbb R^3$that contains the equator of$S^2$. This equatorial map is a$C^1$-approximation to the composition$S^1\xrightarrow{\text{double covering}}S^1\subset\Pi$, so it is an immersion and has an odd number of double points. But then the double point set of$f$cannot be a union of closed curves. So$f$cannot be an immersion. 4 deleted 2390 characters in body It suffices to show that if Let$f:S^2\to\Bbb R^3$is be a self-transverse map (not necessarily an immersion) that is$C^1$-close to$\phi$(and hence transverse to the equatorial plane), then the double point locus$\Delta_f$\phi$. The image of $f$ meets the equatorial plane lies in an odd number of points.

In order to prove this, consider the singular set a tubular neighborhood $\Sigma_f$ formed by unordered pairs S^2\times\Bbb R$of distinct points the image of$S^2$that get identified under$f$. This is always a manifold (as long as \phi$. Consider the composition $f$ \psi:S^2\xrightarrow{f}S^2\times\Bbb R\xrightarrow{\text{projection}}\Bbb R$. It is self-transverse) and if$f$has no triple points then$\Delta_f=\Sigma_f$(but C^1$-close to $\Delta_f$ \phi$, so it is not a manifold in general). Now equivalent to$\Sigma_f$is \phi$ by a subset change of the quotient $\tilde S^2=S^2\times S^2\setminus\Delta_{S^2}$ by the factor exchanging involution $T$, and it lies in coordinates outside a small closed tubular neighborhood $R$ of $\Sigma_\phi$ in $\tilde S^2/T$. Note that $\Sigma_\phi$ is homeomorphic to $S^1\times\Bbb R$, with the two ends approaching two distinct points on the diagonal poles (which are the singular points of $\phi$).

The remainder of the proof can be presented in two ways.

Algebraic proof (sketch).A simple old argument (see for instance Lemma 7 in http://arxiv.org/abs/math.GT/0305158) shows

So we may assume thatthe class of $\Sigma_f$ in $H_1^{lf}(R;\,\Bbb Z/2)$ equals the cap-product$[\Sigma_\phi]\smallfrown w_1(T)$, where $w_1(T)$ is the first Stiefel-Whitney class outside of the 1-dimensional bundle associated with the double covering $\tilde S^2\to\tilde S^2/T$, and the locally-finite homology is defined in a small neighborhood of the usual way using possibly infinite chains $C_\ast^{lf}=Hom(C^\ast_c;\Bbb Z/2)$poles, where $C^\ast_c$ denotes cochains with compact support.(There f$is a little technical problem here since vertical lift of$z^2:S^2\to S^2$is not self-transverse, so to be honest one has to start \phi$ (with a self-transverse respect to the projection $C^1$-approximation of this mapS^2\times\Bbb R\to S^2$). Such an approximation is discussed for instance Then, in Example 6 of the same reference.)If you work it out you'll see that$[\Sigma_f]=[pt\times \Bbb R]\in H_1^{lf}(S^1\times\Bbb R)=H_1^{lf}(R)$particular, so$\Sigma_f$cannot be a compact manifold. Hence$f$cannot be an immersion. Geometric proof.It suffices to show that$\Sigma_f$has an odd number of intersection points with sends the manifold$\Pi_f$formed by unordered pairs of points equator of$S^2$such that at least one point of the pair goes into the equatorial plane under$f$. Note that$\Pi_f\cap\Sigma_f$lies \Pi$ in a small neighborhood $N$ \Bbb R^3$that contains the equator of$\Pi_\phi\cap\Sigma_\phi$in$\bar S^2$. S^2$. This neighborhood is compact (i.e. does not approach the diagonal) since so equatorial map is a $\Pi_\phi\cap\Sigma_\phi$ itself (it is homeomorphic C^1$-approximation to the composition$S^1$). CertainlyS^1\xrightarrow{\text{double covering}}S^1\subset\Pi$, $\#(\Sigma_{f_0}\cap\Pi_{f_0})$ so it is an immersion and has an odd for some $C^1$-approximation $f_0$ number of $\phi$. (You can first approximate the composition $S^1\xrightarrow{2-\text{fold covering}} S^1\subset\Bbb R^2$ by a self-transverse immersion with one double point, and then extend this approximation on the equator to both hemispheres)points.

If $f_1$ is another $C^1$-approximation, But then it is joined to $f_0$ by a self-transverse homotopy $f_t$. The singular the double point set of this homotopy is a bordism between $\Sigma_{f_0}$ and $\Sigma_{f_1}$. Then we also get a bordism between $\Sigma_{f_0}\cap\Pi_{f_0}$ and $\Sigma_{f_1}\cap\Pi_{f_1}$. This bordism is f$cannot be a compact manifold (i.e. does not approach the diagonal) since it is contained in$N$. union of closed curves. So$\#(\Sigma_{f_1}\cap\Pi_{f_1})$must be odd.But as Anton observed it f$ cannot be odd for an immersion, so we get a contradiction.

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The idea of Anton Petrunin can be made into an accurate proof. One does not need $C^1$-closeness C^2$convergence,$C^1$convergence is enough. Note That is, I claim that there is no$C^1$immersion sufficiently$C^1$-close to the composition$\phi:S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$. (By the way, any map$S^2\to\Bbb R^3$is$C^0$-close to a smooth$C^\infty$immersionby , according to the$C^0$-dense$h$-principle (for this it is enough to know and using that$S^2$immerses in$\Bbb R^3$). I will denote the composition$S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$by$\phi$. R^3$.)

It suffices to show that if $f:S^2\to\Bbb R^3$ is a self-transverse map (not necessarily an immersion) that is $C^1$-close to $\phi$ (and hence transverse to the equatorial plane), then the double point locus $\Delta_f$ of $f$ meets the equatorial plane in an odd number of points.

In order to prove this, consider the singular set $\Sigma_f$ formed by unordered pairs of distinct points of $S^2$ that get identified under $f$. This is always a manifold (as long as $f$ is self-transverse) and if $f$ has no triple points then $\Delta_f=\Sigma_f$ (but $\Delta_f$ is not a manifold in general). Now $\Sigma_f$ is a subset of the quotient $\tilde S^2=S^2\times S^2\setminus\Delta_{S^2}$ by the factor exchanging involution $T$, and it lies in a small closed tubular neighborhood $R$ of $\Sigma_\phi$ in $\tilde S^2/T$. Note that $\Sigma_\phi$ is homeomorphic to $S^1\times\Bbb R$, with the two ends approaching two distinct points on the diagonal (the singular points of $\phi$).

The remainder of the proof can be presented in two ways.

Algebraic proof (sketch). A simple old argument (see for instance Lemma 7 in http://arxiv.org/abs/math.GT/0305158) shows that the class of $\Sigma_f$ in $H_1^{lf}(R;\,\Bbb Z/2)$ equals the cap-product $[\Sigma_\phi]\smallfrown w_1(T)$, where $w_1(T)$ is the first Stiefel-Whitney class of the 1-dimensional bundle associated with the double covering $\tilde S^2\to\tilde S^2/T$, and the locally-finite homology is defined in the usual way using possibly infinite chains $C_\ast^{lf}=Hom(C^\ast_c;\Bbb Z/2)$, where $C^\ast_c$ denotes cochains with compact support. (There is a little technical problem here since $z^2:S^2\to S^2$ is not self-transverse, so to be honest one has to start with a self-transverse $C^1$-approximation of this map. Such an approximation is discussed for instance in Example 6 of the same reference.) If you work it out you'll see that $[\Sigma_f]=[pt\times \Bbb R]\in H_1^{lf}(S^1\times\Bbb R)=H_1^{lf}(R)$, so $\Sigma_f$ cannot be a compact manifold. Hence $f$ cannot be an immersion.

Geometric proof. It suffices to show that $\Sigma_f$ has an odd number of intersection points with the manifold $\Pi_f$ formed by unordered pairs of points of $S^2$ such that both go at least one point of the pair goes into the equatorial plane under $f$. Note that $\Pi_f\cap\Sigma_f$ lies in a small neighborhood $N$ of $\Pi_\phi\cap\Sigma_\phi$ in $\bar S^2$. This neighborhood is compact (i.e. does not approach the diagonal) since so is $\Pi_\phi\cap\Sigma_\phi$ itself (it is homeomorphic to $S^1$).

Certainly, $\#(\Sigma_{f_0}\cap\Pi_{f_0})$ is odd for some $C^1$-approximation $f_0$ of $\phi$. (You can first approximate the composition $S^1\xrightarrow{2-\text{fold covering}} S^1\subset\Bbb R^2$ by a self-transverse immersion with one double point, and then extend this approximation on the equator to both hemispheres).

If $f_1$ is another $C^1$-approximation, then it is joined to $f_0$ by a self-transverse homotopy $f_t$. The singular set of this homotopy is a bordism between $\Sigma_{f_0}$ and $\Sigma_{f_1}$. Then we also get a bordism between $\Sigma_{f_0}\cap\Pi_{f_0}$ and $\Sigma_{f_1}\cap\Pi_{f_1}$. This bordism is a compact manifold (i.e. does not approach the diagonal) since it is contained in $N$. So $\#(\Sigma_{f_1}\cap\Pi_{f_1})$ must be odd. But as Anton observed it cannot be odd for an immersion, so we get a contradiction.

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