If $M\subset Y$ is a subspace, and $Y/M$ has finite dimension, then is $M$ complemented?

Question

If $M\subset Y$ is a subspace, and $Y/M$ has finite dimension, then is $M$ complemented? Why?

The motivation for my question comes from in some functional analysis texts stating that $\operatorname{im}(T)$ being closed is redundant for a Fredholm Operator $T:X\rightarrow Y$. The proofs I've seen of these assume there is a closed subspace $N\subset Y$ such that $Y=\operatorname{im}(T)\oplus N$.

https://math.stackexchange.com/questions/3307828/why-is-the-closed-operatornameim-t-in-the-definition-of-fredholm-operator-r — An old man in the sea., Jul 30 '19 at 08:56
https://math.stackexchange.com/questions/3307725/decomposition-y-textimt-oplus-n-for-fredholm-operator-t — An old man in the sea., Jul 30 '19 at 08:56

score 1 · Answer 1 · answered Jul 30 '19 at 15:12

1

If you have a hyperplane that isn't closed, then it can't be complemented (because a complemented hyperplane is closed) but $Y/M $ is finite dimensional.

answered Jul 30 '19 at 15:12

Paul

261

If $M\subset Y$ is a subspace, and $Y/M$ has finite dimension, then is $M$ complemented?

1 Answers1