Answer
H is one-to-one and onto
Work Step by Step
$A\,function\,\,H: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}\times \mathbb{R} \,is\,\,one-to-one\,\Leftrightarrow \\
\forall \,\,x_{1}\,and\,x_{2}\,\,in\,\,\mathbb{R}\times \mathbb{R} \,\,if\,
H(x_{1}) = H(x_{2})\,\,then\,x_{1} = x_{2}.\\
let \,x_{1}=\left ( a,b \right ), \,x_{2}=\left ( c,d \right )\\
if\,\,H(a,b)=H(c,d) \\
\Rightarrow (a+1,2-b)=(c+1,2-d)\\
\Rightarrow a+1=c+1 ,\,\,2-b=2-d \\
\Rightarrow a=c , b=d \\
\Rightarrow (a,b)=(c,d)\\
\because H(a,b)=H(c,d)\Rightarrow (a,b)=(c,d)\\
\therefore H \,\,is\,one-to-one \\
$
$
H:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}\times \mathbb{R} \\
H(x,y)=(x+1,2-y)\,for\,all\,(x,y)\in\mathbb{R}\times \mathbb{R} \\
H: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}\times \mathbb{R} is\,\,onto\,\Leftrightarrow \,\\
\forall y\,in\,\mathbb{R}\times \mathbb{R} ,\exists x \in \mathbb{R}\times \mathbb{R}\,such\,that\, H(x) = y.\\
let\,y=(f,g) ,\,\,\,x=(w,z)\\
suppose\,(w,z)=(f-1,2-g)\\
H(w,z)=((f-1)+1,2-(2-g))=(f,g)\\
H\,\,is\,\,onto \\
$
