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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

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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.

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