I Need to show that $T:V_1\to V_2$ is onto iff $T^*:V_2^*\to V_1^*$ is one one. Here $V_1,V_2$ are vector spaces and $T$ is linear transformation
I do not want to use rank/transpose argument.
Left to Right (The one I completed and think is correct)
Assume $T$ is onto
To show: $T^*$ is $1-1$
TS: $T^*(v_2^*)=T^*(v_2^{'*})\implies v_2^*=v_2^{'*}$
TS: $\forall v_1\in V_1,\,\,\,T^*(v_2^*)(v_1)=T^*(v_2^{'*})(v_1)\implies v_2^*=v_2^{'*}$
TS: $\forall v_1\in V_1,\,\,\,v_2^*(T(v_1))=v_2^{'*}(T(v_1))\implies v_2^*=v_2^{'*}$
which is true since $T$ is onto
However, the right to left part (The one I need a little help with)
Assume $T^*$ is $1-1$
i.e. $\forall v\in V_1,\,\,T^*(v_2^*)(v)=T^*(v_2^{'*})(v)\implies v_2^*=v_2^{'*}$
i.e. $\forall v\in V_1,\,\,v_2^*(T(v))=v_2^{'*}(T(v))\implies v_2^*=v_2^{'*}$
I think this should imply that $T$ is onto because if it wasn't onto that would mean that even if $v_2^*(y)\ne v_2^{'*}(y),$ it is possible that $v_2^*=v_2^{'*}$
However I'm not sure how to frame this argument. Any help is appreciated
if $\;T^*\;$ is $\;1-1\;$, then $\;\ker T^*=\{0\}\;$ , which means
$$T^*f=0\iff f=0\iff \forall x\in V_1\;,\;\; T^*f(x):=f(Tx)=0\iff f=0$$
The very last equality means $\;f(y)=0\;\;\;\forall\,y\in V_2\;$ . If $\;T\;$ isn't onto, there exists $\;0\neq y\in V_2\;$ such that $\;y\notin T(V_1)\;$. Let $\;\{y, y_1,...,y_k\}\;$ be a basis of $\;T_2\;$ ; then we know we can define $\;f\in T^*_2\;$ s.t. $\;fy\neq0\;,\;\;fy_k=0\;\;\;\forall\,k=,1,...,k\;$, and thus for this $\;f\;$ we'll get $\;T^*f=0\;$ , yet it isn't true that $\;f=0\;$ ...!