The isomorphism between $F^B$ and $V$ of dimension $|B|$ $F^B$  is  the  set  of  functions  from  the  set  $B$   to  the  set  $F$  with  finite support . Let $B$  has  cardinality  $\kappa $ . Then to  show  any vector  space $V$ of  dimension  $\kappa$  over $F$  is  isomorphic  to  $F^B$.
If the  number  $\kappa$  is  finite  or  countable  then  I  can  show  the  bijection . For uncountable  if  I  do  the  following , is  it  correct  $?$ 
Let $V_B$  be  the basis  of  $V$  and  $D_B=\{\delta_{b}(x)=\delta_{bx}\}$  be  the  basis  for $F^B$. 
Let $\Psi$  be  the  correspondence   between $F^B$  and  $V$ 
Choose  any $\delta_{b_k}$  from  $F^B$  and  any $v_k$ and  write $$\
\Psi(\delta_{b_k})=v_k$$
Now for  the  sets $D_B\backslash \delta_{b_k}$ and $V_B\backslash v_k$. Similar  arbitrary  choices  make  $\Psi $  into  a  bijection . 
Also  , after  bijection is proved , I have to show linearity  of  the  map . i.e. $$\Psi(\alpha\delta_{b_k}+\beta\delta_{b_j})=\alpha\Psi(\alpha\delta_{b_k})+\beta\Psi(\alpha\delta_{b_j})=\alpha v_k +\beta v_j$$
How  do  I  prove  that $?$
 A: Using your notation, by assumption there is a bijection $f : D_B \to V_B$. We are going to define a linear map $\Psi_f : F^B \to V$. Let $x \in F^B$, then we can express $x$ as a linear combination $x = \sum_{b \in B} x_b \delta_b $ (only finitely many $x_b$ non-zero). Now we set $$\Psi_f(x) = \sum_{b \in B} x_b f(\delta_b).$$ This is well-defined, because the linear combination expressing $x$ in the basis $D_B$ is unique.
This gives a well-defined map $\Psi_f : F^B \to V$. To check linearity, note that for $x,y \in F^B$ and $\alpha,\beta \in F$ writing $x = \sum_{b} x_b \delta_b$, $y = \sum_{b} y_b \delta$, then $\alpha x + \beta y = \sum_{b} (\alpha x_b + \beta y_b) \delta_b$, so that 
$$ \Psi_f(\alpha x + \beta b) = \sum_{b} (\alpha x_b + \beta y_b) f(\delta_b) = \alpha \sum_{b} x_b f(\delta_b) + \beta \sum_{b} y_b f(\delta_b) = \alpha\Psi_f(x) + \beta \Psi_f(y)$$ 
where in the first and third step we use the definition of $\Psi_f$ and in the second we just rewrite the sum.
Now using that $f$ is injective if $\Psi_f(x) = \Psi_f(y)$ then we conclude from $V_B$ being a basis (in particular linear, independent) that $x_b = y_b$ for all $b \in B$. Hence $\Psi_f$ is injective.
Also for $v \in V$ we express $v$ as a linear combination $v = \sum_{b \in B} v_b f(\delta_b)$ because $V_B$ is a basis of $V$ and $f$ is surjective. Now we see that $\Psi_f(\sum_{b} v_b \delta_b) = v$ so $\Psi_f$ is surjective.
Putting all things together, we get that $\Psi_f$ is a linear isomorphism between $F^B$ and $V$.
Note that this shows in fact the following.
If $V$ and $W$ are two vector spaces over $F$ with basis $B_V$ and $B_W$, respectively, then any map $f : B_V \to B_W$ extends to a linear map $\Psi_f : V \to W$ in a unique way. Moreover, $f$ is injective (surjective) if and only if $\Psi_f$ is injective (surjective).
