Why does continuous linear functional $f$ in topological space $V$ have to have an inverse?

Question

Like in the proof for kernel of continuous functional being closed.

If f is continuous then $\ker f=f^{-1}(0)$ ...

But how does one know $f^{-1}$ exists?

http://mathworld.wolfram.com/Preimage.html – Nate Eldredge Jun 05 '18 at 15:56 — Nate Eldredge, Jun 05 '18 at 15:56

score 0 · Answer 1 · answered Jun 05 '18 at 15:45

0

It doesn't have to have an inverse. The notation $f^{-1}(0)$ means the set of all vectors $v$ from the domain such that $f(v)=0$.

answered Jun 05 '18 at 15:45

But isn't the argument that follows " If f is continuous then $\ker f=f^{-1}(0)$ and therefore is closed." based on ${0}$ being closed and then a continuous linear functional maps closed to closed by open mapping theorem? So to me it reads like it utilizes the inverse of $f$. – mavavilj Jun 05 '18 at 15:56
No, the argument doesn't use the open mapping theorem at all. The preimage of closed/open sets under a continuous function is always closed/open (respectively). That is actually the definition of continuity for functions between topological spaces. As ${0}$ is closed and $f$ is continuous, $f^{-1}({0})$ is closed. – Michael L. Jun 05 '18 at 15:59

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