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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.

I think if I can show that $T(B_X)$ contains an open ball then I am done where $B_X$ is the unit ball in $X$. But I am unable to show that. Need some help..

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  1. If $T$ is continuous, then $\ker(T)$ is closed and the quotient map $$ \pi : X \to X/\ker(T) $$ is an open map. Furthermore, $T$ induces an injective map $$ S : X/\ker(T) \to Y $$ Since $Y$ is finite dimensional, so is $X/\ker(T)$, and so $S$ (whose range is $Y$) is now a homeomorphism. In particular, $S$ is an open map, so $$ T = S\circ \pi $$ is also open.

  2. If $\dim(Y) = 1$, then it follows from an earlier question that if $U$ is a non-empty open set, then $T(U) = \mathbb{C}$, so it is, in particular, an open map.

  3. Not sure about the general case (if $T$ is discontinuous and $\dim(Y) > 1$), but perhaps someone else can complete that case (I don't think induction works, but perhaps it could)

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  • $\begingroup$ If T is continuous then it is the open mapping theorem..Only Interesting case is to prove the result when T is discontinuous. $\endgroup$ – Saikat Oct 2 '15 at 9:28
  • $\begingroup$ $X$ need not be complete, so open mapping theorem does not necessarily apply. And even if $X$ is complete, the proof above is far easier because the range is finite dimensional. $\endgroup$ – Prahlad Vaidyanathan Oct 2 '15 at 9:32
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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.

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