Is this map to a finite dimensional topological vector space an open map? Suppose $X$ and $Y$ are topological vector spaces, $\dim (Y)<\infty$, and $f:X\to Y$ is linear and surjective. Prove that $f$ is open and if the null space $N$ of $f$ is closed, then $f$ is also continuous.
The second assertion is easier since the quotient space $X/N$ has the universal mapping property.
 A: Here is a tedious proof of the first assertion:
To show that $f$ is open, it is sufficient to show that the image of a neighborhood of $0 \in X$ contains a neighbourhood of $0  \in Y$.
Let $y_1,...,y_n$ be a basis for $Y$ and let $x_i \in f^{-1} \{ y_i \}$. Note that the $x_i$ are linearly independent. Define $\Lambda : \mathbb{F}^n \to X$ by $\Lambda \alpha = \sum_k \alpha_k x_k$. $\Lambda$ is linear, continuous and $\ker \Lambda = \{ 0\}$.
Let $U$ be a neighborhood of $0 \in X$. By continuity of $\Lambda$, $\Lambda^{-1} U$ is open, hence contains $B(0, \delta) \subset \mathbb{F}^n$ for some $\delta>0$.
Define $\Phi:\mathbb{F}^n \to Y$ by $\Phi = f \circ \Lambda$, and note that $\Phi$ is bijective, hence $\Phi(B(0, \delta))$ is open (and contains $0$). Since $0 \in \Phi(B(0, \delta)) = f (\Lambda (B(0, \delta))) \subset f(U)$, we see that $f(U)$ contains a neighborhood of $0 \in Y$. Hence $f$ is open.
For the second part, let $\pi: X \to X/\ker f$ be the quotient map. Note that $\ker f$ is closed, hence $X/\ker f$ is a topological vector space (in fact this is iff, closedness is needed to ensure that $X/\ker f$ is Hausdorff). Furthermore, $\pi$ is (by definition) continuous. The map $\tilde{f}: X/\ker \to Y$ is a bijection, hence $X/\ker$ is finite dimensional, and so$\tilde{f}$ is continuous. Since $f = \tilde{f}  \circ \pi$, it follows that $f$ is continuous.
Alternatively, one could show that a functional is continuous iff its kernel is closed (Rudin's route involves showing that $\ker f$ closed implies $\ker f$ is not dense, and that the latter implies that $f$ is bounded, and continuity follows from this). Continuity of linear maps between finite dimensional topological vector spaces finishes the proof.
