Possibly a very quick and simple question:
I was reading an elementary textbook when I encountered some confusion about the term small image under some function $f: X \to Y$.
$M$ is an arbitrary subset of $X$. Then the small image of $M$ under $f$ is defined as the set $\left\{y \in Y: f^{-1}(y) \subseteq M\right\}$ denoted with $f^\#(M)$. [...] Thus, $f^\#(M) = Y \setminus f(X \setminus M)$.
I haven't found any reference to the term small image in connection to this topic anywhere else. I would assume that if $f^\#(M)$ is the set of all $y$ values "originating" from $M$ it should be equivalent to $f(M)$. But the author asserts they are not, later stating $f^\#(M) \subseteq f(M)$ (with the condition that all $y$ values are assigned at least one $x$ value) without much further explanation.
The book is quite old, so it might suffer from outdated terminology.
I haven't heard of this terminology, but the assertion $f^\#(M) \subseteq f(M)$ is correct, and $f^\#(M) =f(M)$ if and only if $M$ is saturated, that is, $M=f^{-1}(f(M))$. $$y\in f^\#(M)\implies f^{-1}(y) \subseteq M\implies y\in f(M)$$ The key point is that for $y$ to belong to $f^\#(M)$, the complete preimage of $y$ must lie in $M$, whereas for $y$ to belong to $f(M)$, only one point of the preimage of $y$ needs to lie in $M$.
For example, take the function $f(x)=|x|$, let $M=[0,1]$, then $1\in f(M)$ because $f(1)=1$, but $1\notin f^\#(M)$ since $f^{-1}(1)=\{-1,1\}\not\subseteq M$.