In Nakahara's book "Geometry, Topology, and Physics" there is an exercise that asks to show that if $f$ is a Linear Map between vector spaces: $f: V \rightarrow W$ then it is a tensor of type $(1, 1)$.
The exercise is fairly simple because there isn't a lot of ways to "force" $f$ to fit this definition, clearly for $v \in V$ we have $f(v) \in W$ and hence $f(v)$ is a type $(1,0)$ tensor on $W$. So, by virtue of this we can go on to say that since $f(v)$ acts on a co-vector $\omega \in W^{\ast}$ we have a map that takes pairs of the form $(\omega, f(v))$ to some underlying field $K$ and hence establish that $f$ is a $(1, 1)$ tensor.
My difficulty with this first and foremost, is that $f$ is not guaranteed to be surjective on $W$, since it is only said to be a Linear Map and nothing further. So can it be a proper Tensor on $W$ at all in such a case?
A secondary problem is how contrived it appears to force $f$ to fit this definition, and I am not sure why and where is it useful to do this. I'll be glad if someone can help me to clarify this.
In case of finite dimensions and if you have basis $$V={\rm span}\{b_1,...,b_n\},$$ $$W={\rm span}\{c_1,...,c_m\},$$ then one defines $L_{f\otimes w}:V\to W$ via $$L_{f\otimes w}x=f(x)w,$$
where $V^*={\rm span}\{\beta^1,...,\beta^n\}$ is the dual of $V$ and so $\beta^k(b_j)=\delta^k_j$.
That transformation is linear and the element $f\otimes w$ is in $V^*\otimes W$.
The task of checking that $L:V^*\otimes W \to \hom(V,W)$, mapping $$f\otimes w\mapsto L_{f\otimes w},$$ is a vector spaces isomorphism, would be a routine.
Observation:
The elements $\beta^k\otimes c_j$ serve to generate $V^*\otimes W$.