# Prove that $T$ is bounded if its kernel is closed - Proof Clarification

Let $$T : X \rightarrow Y$$ be a map between two normed spaces, where $$Y$$ is finite dimensional. Let $$U$$ be the kernel of $$T$$. Prove that $$T$$ is bounded if $$U$$ is closed.

If $$\ker(T)$$ is closed then $$X/\ker(T)$$ is a normed vector space. Observe that the map $$\overline{T}:X/\ker(T)\to Y$$ given by $$\overline{T}(x+\ker(T))=T(x)$$ is a well-defined linear map by the first part $$\overline T$$ is continuous since $$X/\ker(T)$$ is a finite dimensional vector space (as it is isomorphic to a subspace of $$Y$$). Let $$\pi: X \to X/\ker(T)$$ denote the quotient map. Note that $$T=\overline{T}\circ \pi$$ hence $$T$$ is continuous since it is a composition of continuous functions.

I have a couple questions.

1) Is it possible to show this if $$T$$ is not linear? (My problem doesn't explicitly give this linearity; however, I kind of think that it is just assumed because we are talking about linear operators.)

2) I don't understand why $$X/ker(T)$$ is finite.

if $$T$$ is not linear the result is false even when $$X=Y=\mathbb R$$. (try to construct a function $$f:\mathbb R \to \mathbb R$$ which vanishes only at $$0$$ but not bounded on $$[-1,1]$$. Since there is no continuity assumption this is very easy.
For 2) define $$S:X|ker(T) \to T(X)$$ by $$S(x+Ker(T))=Tx$$. Verify that this linear map is a bijection. Hence the dimension of $$X|ker(T)$$ is equal to that of $$T(X)$$, hence less than or equal to the dimension of $$Y$$.