Let $Y$ be a finite-dimensional normed space, $X$ a normed space, and $T: X \to Y$ a surjective linear operator. Show that $T$ is an open mapping.

Lemma : Let $T$ is an onto linear map from $X$ to $Y$, where $X,Y$ are n.l.s and $Y$ is finite dimensional. Let $U \subseteq X$ is open such that $T(U)$ has only non zero elements.Then $T(U)$ is open.

Proof : Let dimension of $Y$ be a natural no. k. Let $u \in U$ ,then $Tu \ne 0$. Extend $Tu$ to a basis of $Y$ say $\{Tu =: u_1,u_2, .... ,u_k\}$. Clearly $u \ne 0$. Choose $v_i \in X$, such that $Tv_i = u_i \forall i = 1, ..., k$ and $v_1 = u$. Thus $\{v_1,v_2, .... ,v_k\}$ is a linearly independent set in $X$. Let $ X_u := span \{v_1,...., v_k\} \le X$. Let $T_u$ denote the restriction of $T$ onto $X_u$. Now $T_u$ : $X_u$ $ \rightarrow Y$ is a linear isomorphism (also homeomorphism of topological spaces).Now $U$ $\cap$ $X_u$ is open in $X_u$, therefore $T(U \cap X_u)$ is also open in $Y$. Now $u \in U \cap X_u$ so $Tu \in T(U \cap X_u) \subseteq T(U) \subseteq Y$. Thus $T(U)$ is open in $Y$.

Answer to the above question.

Enough to show $T(B(0,r))$ is open in $Y$ for all $r \gt 0$ (since translations are homeomorphisms). We have assumed the dimension of $Y$ to be a natural no., otherwise the result is vacuous.If $T(B(0,r))$ is $Y$ then we are done, else choose $y \in Y\backslash T(B(0,r))$. Let $t_y$ denote the translation in $Y$ by $y$. Let $x \in X$ such that $Tx = y$. Let $t_x$ denote the translation by $x$ in $X$. (note $y,x$ are non zero vectors) Clearly $T(t_x(B(0,r))=B(x,r)) = t_y(T(B(0,r))) = T(B(0,r)) + y$. We show that the last set do not contain zero : Let $z \in B(0,r)$, then $Tz + y = 0$ implies $T(-z) = y$, therefore $y \in T(B(0,r))$, contradiction (since$-z \in B(0,r)$). Thus the last set is open in $Y$ by the lemma. Now since translations are homeomorphisms, we have $T(B(0,r))$ is open. This completes the proof.