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Let $f:M\to N$ be a function. For all $Y$ subset of $N$, $f(f^{-1}(Y))$ is a subset of $Y$.

When $f$ is surjective, $f(f^{-1}(Y))$ equals $Y$. Is this true or false?

I guess because of $f(f^{-1}(Y)) = \mathrm{id}(Y)$ this statement is true. Can give some one a better explanation?

egreg
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Arji
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    Well, you should remark that surjectivity guarantees us that every element of $Y$ occurs as $f(x)$ for at least one value of $x\in M$. – lulu Oct 24 '16 at 21:35
  • I think you are confusing the function $f^{-1}$ with the symbol $f^{-1}$ in $f^{-1}(Y)$. Although $f$ may not be bijective (and therefore there is no inverse function), we may define what is called the pre-image of a set: so let $f:X\rightarrow Y$ be a function and $B\subset Y$, the pre-image of $B$ is $f^{-1}(B)={x\in X: f(x)\in B}$ – Arthur Oct 24 '16 at 21:37
  • The image of a set $A\subset X$ is $f(A)={f(a): a\in A}$ – Arthur Oct 24 '16 at 21:41
  • @Arthur This is one reason for which I tend to use $f^{\gets}(Y)$ and $f^{\to}(X)$ instead of $f^{-1}(Y)$ and $f(X)$. It avoids confusion, albeit it's not a standard notation. – egreg Oct 24 '16 at 21:51
  • @egreg I really like your notation, I've never seen it in any book, as you said, it is not standard. Did you create it yourself? – Arthur Oct 24 '16 at 21:57
  • @Arthur I'm not sure; I think to have seen something similar somewhere, several years ago. Somebody else might write $f^!$ and $f_!$ or $f^$ and $f_$ (these are indeed functorial). By the way, there are cases when $f(a)$ might be ambiguous, when $a$ is both an element and a subset of the domain of $f$; think to ordinals. – egreg Oct 24 '16 at 22:01
  • http://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages – Asaf Karagila Oct 24 '16 at 22:18

2 Answers2

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Surjective means every element of $N$ has an (or more) inverse image(s) in $M$.

So $f^{-1}(Y)$ has elements of $M$ for which image under $f$ has every element of $Y$. So $f ( f^{-1}(Y) ) =Y$

P Vanchinathan
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With $f^{-1}(Y)$ we denote a subset of $M$, namely $$ f^{-1}(Y)=\{x\in M:f(x)\in Y\} $$ When $X$ is a subset of $M$, we set $$ f(X)=\{y\in N:\text{there exists $x\in X$ such that $f(x)=y$}\} $$

It is not required that $f$ has an inverse for the notation $f^{-1}$ to be used in this context. So, no, your argument is invalid.

In order to prove the first statement, suppose $y\in f(f^{-1}(Y))$. This means there exists $x\in f^{-1}(Y)$ such that $y=f(x)$. By definition, $x\in f^{-1}(Y)$ implies $f(x)\in Y$. Therefore $y=f(x)\in Y$.

Hence we have proved that $f(f^{-1}(Y))\subseteq Y$.

For second statement, you only need to prove the converse inclusion under the assumption that $f$ is surjective.

Suppose $y\in Y$. By surjectivity, $y=f(x)$, for some $x\in M$. But, by definition, $x\in f^{-1}(Y)$, so, again by definition, $y\in f(f^{-1}(Y))$.

egreg
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