The answer is no. (Although the previous example I gave was bad.) Let $C$ be the curve $y^2 = x^2 (x-1)(2-x)$, so $C$ has a smooth component with $1 \leq x \leq 2$, and also a node at $(0,0)$. Let $M$ be the normalization of $C$; explicitly, $$M = \{ (x,y,z) : z^2 = (x-1)(2-x),\ y=xz \}.$$ Then the projection $M \to C$ is $1$-to-$1$ over the smooth component but misses the node. <hr> However, such examples can only miss a set of lower dimension than $M$, and cannot occur if all connected components of the Zariski closure of $f(M)$ have the same dimension! This follows from <cite authors="Bialynicki-Birula, A.; Rosenlicht, M.">_Bialynicki-Birula, A.; Rosenlicht, M._, [**Injective morphisms of real algebraic varieties**](http://dx.doi.org/10.2307/2034464), Proc. Am. Math. Soc. 13, 200-203 (1962). [ZBL0107.14602](https://zbmath.org/?q=an:0107.14602).</cite> At the start of section 2, they prove the following result: >Let $V$ and $W$ be real algebraic sets and let $f: V \to W$ be an injective morphism such that $f(V)$ is Zarsiki-dense in $W$. Then $f(V)$ contains a Zariski-open Zariski-dense subset of $W$. Note that this is very false for $f$ non-injective; consider the map $x \mapsto x^2$ from $\mathbb{R}$ to itself. In our setting, take $V = M$ and let $W$ be the Zariski-closure of $f(M)$. For simplicity, let $M$ be irreducible, so $W$ will be as well. Bialynicki-Birula and Rosenlicht's result shows that $f(M)$ must contain a Zariski-open Zariski-dense subset $U$ of $W$, and thus $W \setminus U$ must be a proper Zariski closed subset of $W$. In particular, $W \setminus U$ must have dimension lower than $\dim W = \dim M$. If we now assume that $W$ is irreducible and all its connected components have the same dimension, then $U$ must be dense in $M$ for the analytic topology. But, also, $f(M)$ is closed in the analytic topology since $M$ is compact. We have shown that $f(M)$ is closed for the analytic topology and contains a dense set (namely $U$) for the analytic topology, so $f(M) = W$.