Proof of a property of fuzzy extension principle I'm looking for a proof of the following property of the fuzzy extension principle:
$$
B \supseteq f(f^{-1}(B))
$$
($f: X\rightarrow Y$ is an arbitrary crisp function, and $B$ is a fuzzy set). I think if f is a function, the two sides would be equal. But I see this property on page 47 of the book by Klir and Yuan.
Thank you.
 A: $f\left(f^{-1}\left(B\right)\right)\subseteq B$ is always true, which
is not the case for $B\subseteq f\left(f^{-1}\left(B\right)\right)$
(see the answer of 5xgum). It is true for each $B$ if and only if
$f$ is surjective.
A: "I think if $f$ is a function, the two sides would be equal"
Not really, look at the function $f:\mathbb R\rightarrow \mathbb R$ defined by $f(x)=x^2$ and $B=[-1,0)$. Since $f^{-1}(B)=\emptyset$, it follows that $f(f^{-1}(B))=\emptyset$.
A: As you said, $f$ is a crisp function, so it will only work with those elements that ARE or ARE NOT in $X$, and the same for $Y$.
Note that $B:Y\to [0,1]$ and $f^{-1}$ will take those $y$ in $Y$ such that $B(y)=1$ (you can call that subset of $Y$, $B^{-1}(\{1\})$) and relate them with some $x$ in $X$ such that $f(x)=y$, if it does exist.
Now, the subset of $X$ made of those $x$'s will be returned to $Y$ with $f$, into a subset of $B^{-1}(\{1\})$, because if some $y$ in $B^{-1}(\{1\})$ actually doesn't have a preimage.
So, $f(f^{-1}(B)) \subset B^{-1}(\{1\}) \subset B$, because in fact it may have existed a $y_0$ in $Y$ such that $B(y_0)=\frac{1}{2}$, for say something, and it wouldn't even be in $B^{-1}(\{1\})$.  
(Sorry for the mecca font; hope this helps you 1 year after you published it)
