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A bit off-topic, but I'd like to mention some big differences between Whitney's $2n$ theorem and his $2n+1$ theorem.

The idea of the $2n+1$ theorem is that "most" smooth maps from a compact smooth $n$-manifold to a $2n+1$-manifold are embeddings -- in particular every map is smoothly homotopic to an embedding, by an arbitrarily short homotopy. This, coupled with the relatively easy result that every continuous map between smooth manifolds is homotopic to a smooth map, implies that every map is homotopic to a smooth embedding. In particular every $n$-manifold embeds in $\mathbb R^{2n+1}$.

The $2n$ theorem has a trickier proof. Step 1, most smooth maps from an $n$-manifold to a $2n$-manifold are immersions (locally embeddings) without triple points or non-transverse double points. Step 2, there is a procedure for eliminating a transverse double point by a homotopy under certain broad hypotheses. Differences: (1) The homotopy is not short. This is not about "most maps" being embeddings. (2) Step 2 fails if $n=2$. That's because you use an embedded $2$-disk in constructing the homotopy, but in the construction of an embedded $2$-disk in a $2n$-manifold won't be had for free as in the $2n+1$ theorem if $2n=4$. (3) Step 2 also requires the choice of some path in the domain and a nullhomotopy of some loop in the codomain, which means that it fails if the given map of manifolds is non-injective on $\pi_0$ or non-surjective on $\pi_1$. It's OK for embedding in $\mathbb R^{2n}$, but there are simple counterexamples in general. Also, for $2$-manifolds in $\mathbb R^4$ you cheat and use the classification of surfaces. For $2$-manifolds in $4$-manifolds there are interesting surprises: not every map $S^2\to \mathbb CP^2$ or $S^2\to S^2\times S^2$ is homotopic to an embedding.

1

A bit off-topic, but I'd like to mention some big differences between Whitney's $2n$ theorem and his $2n+1$ theorem.

The idea of the $2n+1$ theorem is that "most" smooth maps from a compact smooth $n$-manifold to a $2n+1$-manifold are embeddings -- in particular every map is smoothly homotopic to an embedding, by an arbitrarily short homotopy. This, coupled with the relatively easy result that every continuous map between smooth manifolds is homotopic to a smooth map, implies that every map is homotopic to a smooth embedding. In particular every $n$-manifold embeds in $\mathbb R^{2n+1}$.

The $2n$ theorem has a trickier proof. Step 1, most smooth maps from an $n$-manifold to a $2n$-manifold are immersions (locally embeddings) without triple points or non-transverse double points. Step 2, there is a procedure for eliminating a transverse double point by a homotopy under certain broad hypotheses. Differences: (1) The homotopy is not short. This is not about "most maps" being embeddings. (2) Step 2 fails if $n=2$. That's because you use an embedded $2$-disk in constructing the homotopy, but in the construction of an embedded $2$-disk in a $2n$-manifold won't be had for free as in the $2n+1$ theorem if $2n=4$. (3) Step 2 also requires the choice of some path in the domain and a nullhomotopy of some loop in the codomain, which means that it fails if the given map of manifolds is non-injective on $\pi_0$ or non-surjective on $\pi_1$. It's OK for embedding in $\mathbb R^{2n}$, but there are simple counterexamples in general. Also, for $2$-manifolds in $\mathbb R^4$ you cheat and use the classification of surfaces. For $2$-manifolds in $4$-manifolds there are interesting surprises: not every map $S^2\to \mathbb CP^2$ or $S^2\to S^2\times S^2$ is homotopic to an embedding.