Suppose injective function, show there exists a linear map. Let V and W be vector spaces over a field K and T : V → W be a linear map.
Suppose that T is injective. Show that there exists a linear map S : Im T → V such
that S ◦ T(v) = v for all v ∈ V , and T ◦ S(w) = w for all w ∈ Im T.
Since injective, I have that ker(T)=0, then by rank-nullity theorem, I have:
dim(Im(T))=dim(V). Therefore Im(T) is isomorphic to V. Therefore there exists a linear bijective map S: Im(T) → V. 
I don't know how to show that S ◦ T(v) = v for all v ∈ V , and T ◦ S(w) = w for all w ∈ Im T.
Is it implicit from what I have said so far?
Thanks for the help

For $w \in \text{Im } T$, by injectivity we know there is a unique $v \in V$ such that $Tv = w$, set $Sw = v$. Check $S$ is linear, and it pretty clearly inverts $T$ on the image. 
