Submanifolds and real algebraic sets Nash-Tognoli theorem states that any real compact smooth manifold $A$ is diffeomorphic to a smooth real algebraic set.
If $B$ is a compact submanifold of $A$, does this diffeomorphism $f$ induce a diffeomorphism between $B$ and a smooth real algebraic set?
EDIT : As remarked by @KReiser, there are counter-examples. Instead, I am looking for a $C^\infty$ map from A to $\mathbb R^n$, such that $f\vert_B$ is a diffeomorphism from B onto a smooth real algebraic set.
Thanks,
Lisa
 A: Here is a useful fact:
Lemma. Suppose that $M$ is a smooth manifold and $N\subset M$ is a smooth (for simplicity, closed) submanifold. Then every smooth map $f: N\to R^n$ admits a smooth extension $F: M\to R^n$.
Proof. This extension clearly exists locally, near every point $x\in N$. Let $\{U_i: i\in I\}$ denote a locally finite cover of $N$ by open subsets of $M$ such that for each $i\in I$ the map $f|U_i\cap N$ extends to a smooth map $F_i: U_i\to R^n$. Add to this cover the open set $M-N$ and call ${\mathcal V}$ this open cover of $M$ with the index set $J=I\cup \{0\}$. Let $F_0: M-N\to R^n$ be any constant map, say, zero.
Let $\{\eta_i\}$ denote a smooth partition of unity on $M$, corresponding to the open cover ${\mathcal V}$.   Now, take
$$
F:= \sum_{i\in J} \eta_i F_i.
$$
This is the required extension. qed
Now, in your setting, let $f: B\to R^n$ be a diffeomorphism to a smooth algebraic subset of $R^n$. Using lemma, extend $f$ to a smooth map $F: A\to R^n$.
With a bit more work, increasing the dimension, one can get an extension $F: A\to R^m$ which is a smooth embedding. Let me know if you actually want this.

Edit.
THEOREM. If $B$ is a smooth compact submanifold of a compact smooth manifold $A$, then there exists a smooth embedding $\Phi: A\to R^N$ such that $\Phi(B)$ is an algebraic subset of $R^N$.
I will only describe steps of the proof. If you are familiar with differential topology, you should be able to fill in the details. Otherwise, details probably will not be very helpful.

*

*(Nash-Tognoli theorem) There exists a diffeomorphism $f: B\to C\subset R^n$, where $C$ is a smooth algebraic subset of $R^n$.


*Let $\nu(B)$ denote the normal bundle of $B$ in $A$. Then, as every smooth finite-dimensional vector bundle,
$\nu(B)$ admits an additive inverse, i.e. a smooth vector bundle $\xi=(X\to B)$ such that $\nu(B) \oplus \xi$ is a trivial rank $k$ bundle $\epsilon=(E\to B)$. We, thus, by embedding $R^n$ in $R^{n+k}$ (and regarding the standard projection of the latter to the former as a trivial rank $k$ vector bundle $\zeta$), we obtain an embedding of vector bundles $\nu(B)\to \eta$, where $\eta$ is the pull-back of $\zeta$ to $C$.


*Put a  Riemannian metric on $A$. Then the normal exponential map $\nu(B)\to A$ will yield a diffeomorphism of the total space of an open disk subbundle in $\nu(B)$ to an (open) tubular neighborhood $U$ of $B$ in $A$.  Thus, we obtain a smooth embedding
$$
\phi: U=U_0\to R^{n+k}
$$
extending the embedding $f: B\to C$.


*This step of the proof is a bit similar to the proof of the lemma. We cover $A-B$ by finitely many open subsets $U_i, i=1,...,s$,
each diffeomorphic to $R^d$ (where $d$ is the dimension of $A$) and add the open subset $U_0$ to this collection to obtain a finite open cover of $A$. Then using the partition of unity we extend the map $\phi$ to a smooth map $\Phi_0:  A\to R^{n+k}$ which restricts to $f$ on $B$. Of course, this map is far from an embedding.


*For each $i=1,...,s$ we take a diffeomorphism from $U_i$ to the open ball in $R^d$ and extend it to a smooth map $\Phi_i: M\to S^d$, where the latter is the one point compactification of $R^d$, so that $\Phi_i$ is constant on $B$.


*Lastly, take
$$
\Phi=(\Phi_0,...,\Phi_s): M\to R^{n+k} \times S^d \times ... \times S^d. 
$$
This is a variation on Whitney's proof that every smooth compact manifold embeds in some Euclidean space.
Then $\Phi$ is an embedding $A\to R^{n+k+ sd}$ and the image of $B$ is a smooth algebraic subset (the product of $C$ and a singleton). qed
