I'm attempting to solve the following problem:
Let $V$ and $W$ be vector spaces over $\mathbb{F}$. Prove that $$(V\oplus W)^*\cong V^*\oplus W^*$$ by constructing an explicit isomorphism.
I'm pretty sure I know the isomorphism, but I'm struggling to show that it is in fact a bijection. Here's what I have so far:
Note that $$(V\oplus W)^*=\{\varphi|\varphi:V\oplus W\rightarrow\mathbb{F}, \varphi \mbox{ is linear}\},$$ $$V^*\oplus W^*=\{(\psi,\theta)|\psi:V\rightarrow\mathbb{F},\theta:W\rightarrow\mathbb{F},\psi,\theta \mbox{ are linear}\}.$$ Let $\psi\in V^*$ and $\theta\in W^*$ and define $$L_{(\psi,\theta)}:V\oplus W\rightarrow\mathbb{F}$$ by $$L_{(\psi,\theta)}(v,w)=\psi(v)+\theta(w).$$ We will show that $L_{(\psi,\theta)}$ defines an isomorphism from $V^*\oplus W^*$ to $(V\oplus W)^*$. First, observe that since $\psi,\theta$ are linear by definition, it follows that $$L_{(\psi,\theta)}(c_1(v_1+v_2),c_2(w_1+w_2))=c_1\psi(v_1)+c_1\psi(v_2)+c_2\theta(w_1)+c_2\theta(w_2)$$ for all $c_1,c_2\in\mathbb{F}$, $v_1,v_2\in V$, and $w_1,w_2\in W$. Hence, $L_{(\psi,\theta)}$ is linear.
I think the best way to show injectivity is to show that $\ker(L_{(\psi,\theta)})$ is trivial, but it's not immediately obvious to me how to do so. I think surjectivity is shown by just checking the definition, but again this isn't immediately obvious to me.
Expanding on Jose27's comment and my initial attempt:
Note that $$(V\oplus W)^*=\{\varphi|\varphi:V\oplus W\rightarrow\mathbb{F}, \varphi \mbox{ is linear}\},$$ $$V^*\oplus W^*=\{(\psi,\theta)|\psi:V\rightarrow\mathbb{F},\theta:W\rightarrow\mathbb{F},\psi,\theta \mbox{ are linear}\}.$$ Let $\psi\in V^*$ and $\theta\in W^*$ and define $$L_{(\psi,\theta)}:V\oplus W\rightarrow\mathbb{F}$$ by $$L_{(\psi,\theta)}(v,w)=\psi(v)+\theta(w).$$ We will show that $L_{(\psi,\theta)}$ defines an isomorphism from $V^*\oplus W^*$ to $(V\oplus W)^*$ by showing that $L_{(\psi,\theta)}$ is a linear bijection. First, observe that since $\psi,\theta$ are linear by definition, it follows that $$L_{(\psi,\theta)}(c_1(v_1+v_2),c_2(w_1+w_2))=c_1\psi(v_1)+c_1\psi(v_2)+c_2\theta(w_1)+c_2\theta(w_2)$$ for all $c_1,c_2\in\mathbb{F}$, $v_1,v_2\in V$, and $w_1,w_2\in W$. Hence, $L_{(\psi,\theta)}$ is linear. To see that $L_{(\psi,\theta)}$ is injective, suppose $$L_{(\psi_1,\theta_1)}(v,w)=L_{(\psi_2,\theta_2)}(v,w)$$ for some $\psi_1,\psi_2\in V^*$ and $\theta_1,\theta_2\in W^*$, and all $v\in V$ and $w\in W$. Then \begin{align*} L_{(\psi_1,\theta_1)}(0,w)&=L_{(\psi_2,\theta_2)}(0,w)\\ \Rightarrow\psi_1(0)+\theta_1(w)&=\psi_2(0)+\theta_2(w)\\ \Rightarrow\theta_1(w)&=\theta_2(w) \end{align*} for all $w\in W$. Hence $\theta_1=\theta_2$. It is similarly shown that $\psi_1=\psi_2$. Hence $L_{(\psi,\theta)}$ is injective. To see that $L_{(\psi,\theta)}$ is surjective, given $\varphi\in(V\oplus W)^*$, define $$\psi(v):=\varphi(v+0), ~~ \theta(w):=\varphi(0+w),$$ for each $v\in V$, and $w\in W$. It is clear that $\psi$ and $\theta$ are linear, hence, $\psi\in V^*$, $\theta\in W^*$. Therefore, \begin{align*} L_{(\psi,\theta)}(v,w)&=\psi(v)+\theta(w)\\ &=\varphi(v+0)+\varphi(0+w)\\ &=\varphi(v+w) \end{align*} The desired result follows.