Is the image of a $\sigma$-algebra necessarily a $\sigma$-algebra? 
Is the image of a $\sigma$-algebra necessarily a $\sigma$-algebra ?

I know that pre-image is always a $\sigma$-algebra, since $f^{-1}(\bigcap_iB_i)=\bigcap_if^{-1}(B_i)$ and $f^{-1}(\bigcup_iB_i)=\bigcup_if^{-1}(B_i)$. 
I think, that if $f$ is not surjective, then the image is not a $\sigma$-algebra, but what if it's surjective. Do you know other counter-examples.
 A: It is not true, as s.harp pointed out in the comments.
Take $f:[0,2] \to [0,1]$ defined by $f(x) = |1-x|$, then $f$ is
surjective. Take the $\sigma$-algebra ${\cal B}=\{\emptyset, [0,2], [0,1],(1,2] \}$, then the image of ${\cal B}$ is
$\{\emptyset, [0,1],(0,1]\}$ which is not a $\sigma$-algebra.
A: Let $f \,:\, A \to B$ be a function and $\sigma$-Algebra $\mathcal{A}$ on $A$.
Then you're right that technically, if $f$ is not surjective, $$
  f(\mathcal{A}) := \left\{f(M) \subset B \,:\, M \subset A, M \in \mathcal{A}\right\}
$$
isn't a $\sigma$-Algebra over $B$ because for all $X \in f(\mathcal{A})$ you have $X \subset f(A)$ and thus $B \notin f(\mathcal{A})$.
But one could say that you simply picked a bad codomain for $f$ in this case - if you re-define $f$ as $f \,:\, A \to f(A)$, i.e. pick a smaller codomain to make $f$ surjective, than that problem goes away.
If $f$ is injective, it's clear that $f(\mathcal{A})$ is always a $\sigma$-Algebra over $f(A)$, because $f$ simply amounts to a renaming of the elements of $A$ then. That leaves the question of whether that same is true for a non-injective $f$. My (not very educated) guess is that you'll still get a $\sigma$-Algebra over $f(A)$ even then, but I don't have a good argument for that at hand...
A: Here is a smallest counter-example that $f(S)$ is not a $\sigma$-algebra.
Define
$$
X:=\{a,b,c\}, \ Y:=\{1,2\},
$$
$$
S:=\{ \emptyset, X, \{a\}, \{b,c\}\},
$$
and
$$
f(x) =
\begin{cases} 1 & \text{ if } x\ne c\\
2 & \text{ if } x=c.
\end{cases}
$$
Then clearly $S$ is a $\sigma$-algebra on $X$, $f:X\to Y$ is surjective, but
$$
f(S)=\{ \emptyset,  \{1\}, \{1,2\}\}
$$
is not a $\sigma$-algebra over $f(X)=Y=\{1,2\}$.
The example is minimal in the following sense: If $Y$ is a singleton, then $f(S)$ is trivially a $\sigma$-algebra for any $f:X\to Y$. If $f$ is bijective, then $f(S)$ is a $\sigma$-algebra as well. So $Y$ has to consist of at least $2$ elements, $X$ has at least $3$ elements. Also $f(S)$ is the smallest non-trivial $\sigma$-algebra possible.
