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If I have $X,Y$ Banach spaces and $T : X \to Y$ a linear and surjective map then $T$ is a open map. I want to prove that if $(y_n)$ is a bounded sequence on $Y$ then there exists a bounded sequence $(x_n)$ on $X$ such that $T(x_n) = y_n$. I want this because I am trying to prove that if $y_n \to 0$ then there exists a sequence $(x_n)$ in $X$ that converges to $0$ and $T(x_n) = y_n.$

Since $T$ is surjective, clearly there exists $(x_n) \in X$ such that $T(x_n) = y_n$. This is easy. Once $(y_n)$ is bounded it lies on a ball for a radius big enough. So, how can I find a ball in $X$ such that $(x_n)$ lies in a ball? The radius of such ball must be related with the radius of the ball containing $(y_n)$ somehow. How can I do this? It is obviously open map theorem, but how?

Nate Eldredge
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L.F. Cavenaghi
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    For your original question, consider the space $X/\ker T$ and the operator $A : X/\ker T\to Y$ mapping $x+\ker T$ to $Tx$. Then $A$ is bijective... – Friedrich Philipp Apr 01 '17 at 22:49

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Hint: Let $B$ be the open unit ball of $X$. By the open mapping theorem, $T(B)$ contains an open ball centered at the origin of $Y$. By scaling, you can find a sufficiently large ball $B'$ in $X$ such that $T(B')$ contains a ball which contains all the $y_n$. You can then choose all the $x_n$ inside $B'$.

Nate Eldredge
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