Let $\sigma :\{1,2,3,4,5\} \rightarrow \{1,2,3,4,5\}$ be permutations such that $ \sigma^{-1}(j) \leq \sigma(j)~\forall j, 1 \leq j \leq 5$. Which of the following are true?


*

*$\sigma \circ \sigma(j)=j~\forall j, 1 \leq j \leq 5$.

*$\sigma^{-1}(j)=\sigma(j)~\forall j, 1 \leq j \leq 5$.

*The set $ \{k: \sigma(k) \neq k \}$ has even number of elements.

*The set $\{k:\sigma(k) =k \}$ has an odd number of elements . 
Can someone tell me to solve it in $1$ or $2$ min....trick..or some concept behind it?
 A: $\sigma^{-1}(j)\leq\sigma(j)\forall 1\leq j \leq 5$
since $\sigma$ is a permutation,
$j\leq\sigma^2(j)\forall 1\leq j\leq 5$
since $\sigma^2$ is also a permutation, we have $j=\sigma^2(j)\forall 1\leq j\leq 5$
So the longest cycle in $\sigma$ is at most of length $2$.
That is, an element is either mapped to itself, or it is 'paired up' with another element which mapped back to itself. ($\sigma(j)=j$ or $\exists i$ s.t. $\sigma(j)=i$ and $\sigma(i)=j$)
So all of 1,2,3,4 are true.
A: For $k=1,\ldots 5$ denote
$$
i_k = \sigma(k),\; k = \sigma^{-1}(i_k)
$$
Then
$$
k=\sigma^{-1}(i_k) \le \sigma(i_k)
$$
That's why $\sigma(i_5) = 5, \sigma(i_4) = 4, \ldots \sigma(i_1) = 1$.
Since $\sigma(i_k) =  k$ and $\sigma^{-1}(i_k) = k$ we get $$\sigma = \sigma^{-1}$$
Next steps must be obvious :)
A: In general: If $\tau$ and $\sigma$ are permutations on $\left\{ 1,\dots,n\right\} $
with $\tau\left(j\right)\leq\sigma\left(j\right)$ for each $j\in\left\{ 1,\dots,n\right\} $
then with induction it can be shown that $\sigma^{-1}\left(i\right)=\tau^{-1}\left(i\right)$
for $i=1,\dots,n$ or equivalently $\tau=\sigma$.
Base case: If $\sigma\left(k\right)=1$ then also $\tau\left(k\right)=1$
so $\sigma^{-1}\left(1\right)=\tau^{-1}\left(1\right)$.
Suppose it is true for $i=1,\dots,m$ and $\sigma^{-1}\left(m+1\right)=r$
or equivalently $\sigma\left(r\right)=m+1$. Then $\tau\left(r\right)\leq m+1$
but $\tau\left(r\right)=s\leq m$ leads to a contradiction: $r=\tau^{-1}\left(s\right)=\sigma^{-1}\left(s\right)\neq\sigma^{-1}\left(m+1\right)$. 
So in your question we have: $$\sigma^{-1}=\sigma$$
