Prove that if U,W,V are vector spaces above field F, then:
Hom(U,VxW) ~= Hom(U,V) x Hom(U,W)
("~=" - almost equals )
Thank you!
first of all "~=" doesn't mean "almost equals"! it's $\displaystyle \cong,$ which means "isomorphic as vector spaces". anyway, let $\displaystyle \pi_1: V \times W \longrightarrow V, \ \pi_2 : V \times W \longrightarrow W$ be the projection maps, i.e.
$\displaystyle \pi_1(v,w)=v$ and $\displaystyle \pi_2(v,w)=w,$ for all $\displaystyle v \in V, \ w \in W.$ now define $\displaystyle \varphi: \text{Hom}(U, V \times W) \longrightarrow \text{Hom}(U,V) \times \text{Hom}(U,W)$ by $\displaystyle \varphi(f)=(\pi_1f, \pi_2f)$ and show that $\displaystyle \varphi$ is an isomorphism.
you're not doing anything but yelling for more help! by giving you the map $\displaystyle \varphi$ i've already done most of the work for you. the rest is easy. show that $\displaystyle \varphi$ is well-defined, linear, one-to-one and onto.
can't you do any of these? if you want me to write all details for you, that is not going to happen. sorry!