$X$ is a nonempty compact manifold and $Y$ is a connected non-compact manifold, and $f: X\rightarrow Y$ is a smooth function. Need to prove that $f$ has a critical point.
My attempt: Let $f$ does not have a critical point, then all points $y\in Y$ are regular. This means $\forall y\in Y$, $df_x$ is surjective at every point $x\in X$ s.t. $f(x) = y$. I know $f(X)$ is compact, since f is continuous, and connected, since $Y$ is connected, but am not sure how to arrive at a contradiction that $Y$ is non-compact given that $df_x$ is surjective. Thanks and appreciate a hint.
Best Answer
The result is only true under the assumption that $\dim(X)=\dim(Y)$ otherwise the inclusion of a circle in $\mathbb{R}^{2}$ is a counter-example.
If $\dim(X)=\dim(Y)$, assume for a contradiction that $f$ has no critical points. Also assume for simplicity that $X$ is connected (otherwise we can easily apply this argument to each of the connected components of $X$, which all have to be compact by the compactness of $X$).
The non-existence of critical points is equivalent to $f$ being a local diffeomorphism by the inverse function theorem. This implies that $f(X)$ is open in Y, furthermore it is compact as you mentioned which in turn implies that it is closed since $Y$ is metrizable. Hence $f(X)$ is connected, and since $Y$ is connected, we must have $f(X)=Y$, but $Y$ is assumed non-compact hence we obtain the required contradiction.