Solution to operator equation, surjectivity

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Suppose $T:V\to W$, where $V,W$ are banach spaces and $Tf = k$ (for instance $T$ might be an integral operator). They say that the equation has solution when $T$ is injective and so $T^{-1}$ exists. Why the surjectivity is neglected in the litterature? Is it related to the properties of infinite-dimensional spaces?

In the finite dimensional case of a function $f:R\to R$, the bijectivity is required for the existence of $f^{-1}$.

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If $T$ is injective, the solution may not exist for all $k\in W$. However, if the solution exists, it is unique. As a result, injectivity of $T$ assures the uniqueness (but not existence), whereas the surjectivity of $T$ assures the existence: for all $k\in W$ the solution $Tf = k$ exists (but may not be unique).