# Given a linear map $T$, why need it be injective in order to guarantee existence of $S$ with $ST=\operatorname{id}_V$

This is from Axler's Linear Algebra Done Right: Chapter 3 Question 14:

Suppose that $W$ is finite dimensional and $T \in L(V,W)$ Prove that if $T$ is injective, then there exists $S \in L(W,V)$ such that $ST$ is the identity map on $V$?

I do not understand why $T$ has to be injective?

For example, why can't just define $S\in L(W,V)$ such that $S(Tv) = v$

and then $(ST)v = S(Tv) = v$?

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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. –  Julian Kuelshammer Nov 24 '12 at 21:37
ok thanks, ill keep that in mind in the future –  mathnoob Nov 24 '12 at 21:38

If $T$ isn't injective, then there are $u,v\in V$ with $u\ne v$ such that $Tu=Tv$. Then $S(Tu)=S(Tv)$, so you can't have $S(Tu)=u$ and $S(Tv)=v$, since $u\ne v$.
If $T$ is not injective, then we can have $Tv_1 = w = Tv_2$, where $v_1 \neq v_2$ and hence $S$ is not well-defined. For such a map, how would you define $Sw$? If you want to define $S(Tv) = v$, then you have the following inconsistency. $$Sw = S(Tv_1) = v_1$$ and also $$Sw = S(Tv_2) = v_2$$