I am having trouble with the following qualifying exam problem.
Suppose $f: X \rightarrow Y$ is a smooth immersion between smooth manifolds of the same dimension. Given that $X$ is closed and $Y$ is connected, prove that $Y$ is also closed.
I have concluded that $f$ must therefore be a local diffeomorphism and that therefore an open mapping, but I am not sure where to go from there.
Since $f$ is a smooth immersion and since $X$ and $Y$ have the same dimension, by inverse function theorem, $f$ is a local diffeomorphism, and hence an opening mapping. In particular, $f(X)$ is open in $Y$. Since $X$ is compact and since $f$ is continuous, $f(X)$ is also compact, and in particular $f(X)$ is closed in $Y$. Therefore, $f(X)$ is a both open and closed nonempty subset of $Y$, so due to the connectedness of $Y$, $f(X)=Y$. It follows that, (i) $Y=f(X)$ is compact; (ii) since $X$ has no boundary point and since $f$ is a local diffeomorphism, $Y=f(X)$ also has no boundary point.