Answer
a-S is not one-to-one
b- S is not onto
Work Step by Step
$S:\mathbb{Z}^{+}\rightarrow\mathbb{Z}^{+} \\
S(n)=sum\,of\,positive\,divisors\,of\,n \\
S(10)=1+2+5=8 \,\,\,,S(7)=1+7=8 \\ 10\neq 7 \,\,\,but\,\,S(10)=S(7)\\
so\,\,S(n)\,is\,not\,one-to-one \\
S: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+} is\,\,not\,\,onto\,\Leftrightarrow \,\\
\exists y\,in\,\mathbb{Z}^{+} such\,that\,\forall x \in \mathbb{Z}^{+}, S(x) \neq y.\\
since \,2\in \mathbb{Z}^{+} \,and\,there\,is\,no\,x\in \mathbb{Z}^{+} such\,that\,S(x)=2\\
(because\,S(1)=1\,\,and\,\,S(2)=1+2=3)\\
S \,\, is\,not\,onto
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