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Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic.

Why do we thus require the range to be closed?
Or is a subspace with finite codimension necessarily closed?

Moreover, what could happen if the range wouldn't be closed but would be of finite codimension though?

I'm thinking of some example like: $l^p_0\subsetneq l^p,1\leq p\leq\infty$

C-star-W-star
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2 Answers2

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A bounded operator with finite cokernel has closed range (see Abramovich and Aliprantis, An Invitation to Operator Theory, Corollary 2.17).

C-star-W-star
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You need the range to be closed in order to make sense of the cokernel. If $T:X \to Y$ then $$ \text{coker}(T) = Y / \text{range}(T).$$ In order for $Y / \text{range}(T)$ to be a Banach space you require $\text{range}(T)$ to be closed.

Zorngo
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  • Why?! The construction of quotient spaces is firstly purely algebraic as far as I know... – C-star-W-star Jun 02 '14 at 12:19
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    You need $\text{range}(T)$ closed so that $Y/range(T) $ is a Banach space in the norm $$|x + \text{range}(T)| = \inf { |y|_Y : y \in x + \text{range}(T) }.$$ If $\text{range}(T)$ isn't closed, then the quotient norm isn't positive definite. Sorry, I should have been more careful. – Zorngo Jun 02 '14 at 12:27
  • Please read my question: the quotient space construction works for any subspace. So the index would perfectly make sense for any subspace not necessarily closed - so still the question why closeness apart from getting a normed space – C-star-W-star Jun 02 '14 at 15:21