Proof of Invertibility: Linear Map's Surjectivity and Injectivity Condition

[maNoFchangE]

I am trying to understand the following basic proposition about invertibility: a linear map is invertible if and only if it is injective and surjective.
Now suppose $T$ is a linear map $T:V\rightarrow W$. The book I read goes the following way in proving the proposition in the direction when the surjectivity and injectivity act as the condition. Suppose $T$ is injective and surjective and a vector $w \in W$. Then define $Sw$ to be the unique element of V such that $TSw = w$. Therefore $TS$ is an identity transformation in $W$.
Now, I understand that $Sw$ is unique because $T$ is injective, but I don't know how the surjectivity contributes to guarantee that $Sw$ which satisfies $TSw = w$ does exist.

 

[Samy_A]

I am trying to understand the following basic proposition about invertibility: a linear map is invertible if and only if it is injective and surjective.
Now suppose $T$ is a linear map $T:V\rightarrow W$. The book I read goes the following way in proving the proposition in the direction when the surjectivity and injectivity act as the condition. Suppose $T$ is injective and surjective and a vector $w \in W$. Then define $Sw$ to be the unique element of V such that $TSw = w$. Therefore $TS$ is an identity transformation in $W$.
Now, I understand that $Sw$ is unique because $T$ is injective, but I don't know how the surjectivity contributes to guarantee that $Sw$ which satisfies $TSw = w$ does exist.

$T$ being surjective implies that there is a $v \in V$ satisfying $Tv=w$. That $v$ is $Sw$.
If $T$ is not surjective you can't be sure that there will be a $Sw$ satisfying $TSw = w$.

 

[maNoFchangE]

Sorry I am not getting your explanation. If $T$ is surjective then $\textrm{range}(T) = W$, isn't it. How can this information be used to conclude that there is $v$ in the domain space $V$ which satisfies $Tv=w$, while the surjectivity of $T$ concerns the range space $W$ not the domain space $V$?

 

[Samy_A]

Sorry I am not getting your explanation. If $T$ is surjective then $\textrm{range}(T) = W$, isn't it. How can this information be used to conclude that there is $v$ in the domain space $V$ which satisfies $Tv=w$, while the surjectivity of $T$ concerns the range space $W$ not the domain space $V$?

Correct, surjectivity means $T(V)=W$, that every element of $W$ lies in the range of $T$.

That means that for every $w \in W$ there is a $v \in V$ satisfying $Tv=w$.

 

[HallsofIvy]

"Surjective" says that "for any w in W, there exist at least one v in V such that T(v)= w. "Injective says that there is not more than one such v. If T is both "surjective" and "injective" then there exist exactly one such v and that is, by definition, T^-1(w).

 

[maNoFchangE]

"Surjective" says that "for any w in W, there exist at least one v in V such that T(v)= w. "Injective says that there is not more than one such v. If T is both "surjective" and "injective" then there exist exactly one such v and that is, by definition, T^-1(w).

Hi HallsofIvy, thank you.
My background is actually physics and I am just near the beginning of thinking in an abstract manner. In this state of mine, rearrangement and choice of words to translate abstract mathematical line of reasoning really helps me.