Homomorphism space/Linear subspaces/Fact/Proof

(1). It is clear that we have a linear subspace. In order to prove the statement about the dimension, let ${\displaystyle {}U'}$ be an direct complement of ${\displaystyle {}U}$ in ${\displaystyle {}V}$, so that

${\displaystyle {}V=U\oplus U'\,.}$

Let ${\displaystyle {}u_{1},\ldots ,u_{r}}$ be a basis of ${\displaystyle {}U'}$. Every linear mapping from ${\displaystyle {}S}$ maps ${\displaystyle {}U}$ to ${\displaystyle {}0}$, and on ${\displaystyle {}U'}$ (or on a basis thereof) we have free choice. Therefore

${\displaystyle {}S\cong \operatorname {Hom} _{K}{\left(U',W\right)}\,}$

and the statement follows from fact.

(2). It is clear that we have a linear subspace. The natural mapping

${\displaystyle \operatorname {Hom} _{K}{\left(V,U\right)}\longrightarrow \operatorname {Hom} _{K}{\left(V,W\right)}}$

of fact  (2) is injective in this case. Therefore,

${\displaystyle {}T\cong \operatorname {Hom} _{K}{\left(V,U\right)}\,.}$

(3). It is clear that we have a linear subspace. In the finite-dimensional case, let

${\displaystyle {}V=V_{1}\oplus V_{2}\,}$

be a direct sum decomposition. Due to fact, we have

${\displaystyle {}\operatorname {Hom} _{K}{\left(V,W\right)}=\operatorname {Hom} _{K}{\left(V_{1},W\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,}$

and

${\displaystyle {}H=\operatorname {Hom} _{K}{\left(V_{1},W_{1}\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,.}$

Therefore, the dimension equals

${\displaystyle \dim _{K}{\left(V_{1}\right)}\cdot \dim _{K}{\left(W_{1}\right)}+\dim _{K}{\left(V_{2}\right)}\cdot \dim _{K}{\left(W\right)}=\dim _{K}{\left(V_{1}\right)}\dim _{K}{\left(W_{1}\right)}+{\left(\dim _{K}{\left(V\right)}-\dim _{K}{\left(V_{1}\right)}\right)}\dim _{K}{\left(W\right)}\,.}$

(4). Setting

${\displaystyle {}L_{i}={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{i})\subseteq W_{i}\right\}}\,,}$

we have ${\displaystyle {}L=\bigcap _{i=1}^{k}L_{i}}$. Hence, (4) follows from (3).