$\newcommand{\Hom}{\text{Hom}}$ I want to prove the following
If $T\in\Hom_{F}(V,F)$ is non-zero, then $(V/\ker T)\cong F$.
My Attempt:
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I wanted to show $\overline{T}$ in the above diagram, given by $\overline{T}(\overline{\alpha}):=T(\alpha)$, is an isomorphism. Here, $\Pi(\alpha):=\overline{\alpha}$.
I have been able to show $\overline{T}$ is injective. This is because if $\overline{\beta}\in\ker\overline{T}$, then since the above diagram is commutative, this implies $$\beta\in\ker T=\ker\Pi\implies\Pi(\beta)=0\implies\overline{\alpha}=\bar0.$$ So, $\overline{T}$ is injective. But I cannot show $\overline{T}$ is also surjective as that would require, by the First Isomorphism Theorem, that $\operatorname{Im}T=F$. So, how do I show that $\overline{T}$ is surjective ?
$T: V\to F$ linear map and $T\neq 0$ implies , $\exists k\neq 0\in R(T) $.
As $R(T) $ is non trivial subspace of $F$ and $\dim(F) =1$ . Hence $R(T) =F$
Now for any $k\in F, \exists v\in V$ such that $Tv=k$ .
Hence $\overline{T}(v+\ker(T)) =T(v) =k$
Hence $\overline{T}$ is surjective.