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In topology, a function is continuous if inverse of every open set is open. But for the inverse to be well-defined the function should be bijective. For example consider the projection map. It is not injective: $\pi_1(x,y_1)=\pi_1(x,y_2)=x$ with $y_1\neq y_2$. But still it continuous and continuity is proved using the inverse function.

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    $\begingroup$ This is an unfortunate, and even sadder not unique, case of lousy notation and/or naming. Hayden's answer covers this. $\endgroup$
    – Timbuc
    Oct 23, 2014 at 17:19
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    $\begingroup$ It is "inverse image" (also called pre-image) of an open set, not inverse of an open set. And this does not involve any inverse function. $\endgroup$ Oct 23, 2014 at 18:34

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We say that $f:X\rightarrow Y$ is a continuous map if given any open set $U\subset Y$, then $f^{-1}[U]$ is an open set in $X$. This isn't the inverse map we are using, but a related notion called the "pre-image":

Let $f:X\rightarrow Y$ be a function, and for $B\subset Y$, define the pre-image of $B$ to be $$f^{-1}[B]=\{x\in X \mid f(x) \in B\}$$ This definition does not require $f$ to be bijective at all, and it is what is being denoted whenever $f^{-1}$ is mentioned.

(It is also common to use parentheses rather than square brackets, but I've always felt that using the latter is better for being clear about what is meant.)

Worth mentioning is that this induces a map $f^{-1}[\cdot]:\mathcal{P}(Y)\rightarrow \mathcal{P}(X)$ given by sending $B\subset Y$ to $f^{-1}[B]$. There is also the related map $f[\cdot]:\mathcal{P}(X)\rightarrow \mathcal{P}(Y)$ sending $A\subset X$ to $$f[A]:=\{b\in Y \mid \text{there exists $x\in A$ such that $f(x)=b$}\}$$ This also induces a map $f[\cdot]:\mathcal{P}(X)\rightarrow \mathcal{P}(Y)$ sending $A$ to $f[A]$.

However, in general $f^{-1}[\cdot]$ and $f[\cdot]$ are not inverses! In fact, they are inverses if and only if $f$ has an inverse, for we can consider $f[f^{-1}[\{y\}]]$ and $f^{-1}[f[\{x\}]]$ for $x\in X$ and $y\in Y$. Injectivity of $f$ provides the identity $$f^{-1}[f[A]]=A$$ and surjectivity of $f$ provides the identity $$f[f^{-1}[B]]=B.$$

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  • $\begingroup$ So here we can't use $f(f^{-1}(A))=f^{-1}(f(A))=A$? $\endgroup$
    – Mathgrad
    Oct 23, 2014 at 17:04
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    $\begingroup$ @MathewGeorge Nope, that doesn't hold in general. Take for example the constant map $f_{y_0}:X\rightarrow Y$ defined by $f_{y_0}(x)=y_0$ for every $x\in X$. If $Y\neq \{y_0\}$, then $f[f^{-1}[Y\setminus\{y_0\}]]=f[\emptyset]=\emptyset\neq Y\setminus\{0\}$ $\endgroup$
    – Hayden
    Oct 23, 2014 at 17:07
  • $\begingroup$ @MathewGeorge The second identity might also fail; take the same map as in Hayden's example, then $f^{-1}[f[{x}]]=f^{-1}[{y_0}]=X$ for any point $x\in X$. $\endgroup$
    – mdp
    Oct 23, 2014 at 17:12
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    $\begingroup$ This is why it's important to get used to the idea of a function as a relation: there are aspects of the function as a relation that exist and are true regardless of whether or not the set of ordered pairs really IS a function or not. All relations have inverse images. In this case,we really should call it the "inverse image" when talking to beginners to avoid such notational confusion. $\endgroup$ Oct 23, 2014 at 17:32
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    $\begingroup$ You do have that $f[f^{-1}[B]]\subseteq B$ and $A\subseteq f^{-1}[f[A]]$ $\endgroup$ Oct 24, 2014 at 2:10
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The inverse-image of a set $S$ under a function $f$ is the set $\{x : f(x)\in S\}$. This exists regardless of whether the inverse function $f^{-1}$ exists.

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  • $\begingroup$ Thanks. I have edited the question. $\endgroup$
    – Mathgrad
    Oct 23, 2014 at 17:01
  • $\begingroup$ @MathewGeorge: The history does not show that the question was edited at all; it is not clear which change you mean. $\endgroup$ Oct 23, 2014 at 18:35
  • $\begingroup$ @MarcvanLeeuwen : The question was edited, changing "a" to "every". $\endgroup$ Oct 23, 2014 at 19:59

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