I'm doing a homework problem out of Rudin's $\textit{Functional Analysis}$ which is basically a proof of which I have completed some of it, but I'm not sure about the rest of it. Without further ado, here it is:
Suppose $X, Y$ are topological vector spaces, $\Lambda: X \rightarrow Y$ is a linear map, $N$ is a closed subspace of $X$, $\pi: X \rightarrow X/N$ is the quotient map, and $\Lambda x=0$ $\forall x \in N$.
We are asked to prove the following:
$(1)$ There is a unique $f: X/N \rightarrow Y$ such that $\Lambda x = f(\pi(x))$ $\forall x \in X$.
$(2)$ The map $f$ is linear.
$(3)$ $\Lambda$ is continuous iff $f$ is continuous.
$(4)$ $\Lambda$ is an open map iff $f$ is an open map.
So far, I was able to prove $(3)$. To prove $(3)$ since $N$ is closed, by Theorem $1.41$(a) on p. $29$ of Rudin, we have that the quotient map $\pi$ is continuous. Also, the composition of continuous maps is continuous, so if $f$ is continuous, then so is $\Lambda = f \circ \pi$. The converse holds since $\pi$ is an open map by Theorem $1.41$(a). What I'm not so sure about is the proof for $(1)$, $(2)$, and half of the proof for $(4)$. One direction is clear for $(4)$ in that since $\pi$ is an open map and the composition of open maps is an open map, if $f$ is an open map, then so is $\Lambda$. It is the other direction which is given me some trouble, or maybe it might be something obvious which I'm just missing. I would appreciate some guidance for those parts that I haven't been able to prove yet.
Well, since $N$ is in the kernel of of $\Lambda$, the first isomorphism theorem guarantees the existence and unicity of $f$. It also guarantees that $f$ is linear.
Now, for (4), if $\Lambda$ is open and $U$ is open in $X/N$, then by definition of the quotient topology, $\pi^{-1}(U)$ is open in $X$, so that $\Lambda(\pi^{-1}(U))$ is open in $Y$. But $\Lambda(\pi^{-1}(U)) = f(\pi(\pi^{-1}(U)) = f(U)$. So $f$ is open.