Let $V,W$ vectorial spaces. Prove $(V \times W)^*$ is isomorphic to $V^* \times W^*$

Question

Let $V,W$ vectorial spaces. Prove $(V \times W)^*$ is isomorphic to $V^* \times W^*$

Note: $V^*,W^*$ is a dual space.

My work:

Let $T:(V\times W)^*\rightarrow V^*\times W^*$ defined by $T(f,g)=f(v)+g(w)$ with $v,w\in V,W$ and $f\in V^*$, $g\in W^*$
Is easy prove $T$ is lineal.

Go to prove T is injective.

Suppose $T(f,g)=0$ then, $f(v)+g(w)=0\forall v,w\in V,W$
This implies $f(v)=-g(w) \forall v,w\in V,W$
In particular for $w=0$ we have: $f(v)=-g(0)$

I'm stuck here. Can someone help me?

Answer 1

I think you have to define on the other way: Let $v^{\ast},w^{\ast}\in V^{\ast}\times W^{\ast}$, define $\varphi(v^{\ast},w^{\ast})(v,w)=v^{\ast}(v)+w^{\ast}(w)$. Injectivity: Suppose for any $(v,w)\in V\times W$, $\varphi(v^{\ast},w^{\ast})(v,w)=0$, then in particular, realize to $(v,0)$, then $v^{\ast}(v)=0$, this is true for any $v$, so $v^{\ast}=0$, similar to $w^{\ast}$. Surjectivity: For $u\in(V\times W)^{\ast}$, define $v^{\ast}=u(v,0)$ and $w^{\ast}(w)=u(0,w)$, then $\varphi(v^{\ast},w^{\ast})=u$.