I was looking at the answer to this question about inclusion maps and continuity.

A function is continuous if and only if its composition with the inclusion map is continuous.

The answer states that

$\iota^{-1}(U) = U \cap A$

However, how is this defined for points that are not part of $A$. In other words, it doesn't seem like you can take a function $f(A) \to X$ and invert it from $f^{-1}(X) \to A$ if the original function's range was not all of $X$.

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    $\begingroup$ That’s a preimage, not a forward application of a function. $\endgroup$ – Randall Dec 24 '17 at 18:46
  • $\begingroup$ The original question stated "for every open set $U$ in $X$, $\iota^{-1}(U)=U \cap A$". I don't understand your preimage comment in this context. $\endgroup$ – Jeff Dec 24 '17 at 20:53
  • $\begingroup$ $U \cap A$ is precisely the preimage of $U$ under inclusion, which is what their notation means. $\endgroup$ – Randall Dec 24 '17 at 21:02

For any function $f: X \to Y$ and any $B \subset Y$ the set $f^{-1}[B]$ (sometimes denoted $f^{-1}(B)$ in some texts), the preimage of $B$ under $f$, means by definition the set of points in $X$ with values in $B$, so $$f^{-1}[B] = \{x \in X: f(x) \in B\}$$

You should already know it, as continuity between topological spaces is defined as the fact that for any open subset $U$ of $Y$, the set $f^{-1}[U]$ is open in $X$. It should not be confused with a function image.

When we have the inclusion $i : X \to Y$ for a situation $X \subseteq Y$, so $i(x) = x$, and we have a subset $U \subseteq Y$, then $$\{x \in X: i(x) \in U\} = \{x \in X: x \in U\} = X \cap U$$ and so if we need $i$ to be continuous this means that for every open subset $U$ of $Y$, the intersection $X \cap U$ must be open in $X$. Which is exactly the definition of the subspace topology, which is chosen minimally so that $i$ is continuous.

  • $\begingroup$ This clarifies things for me, I think. My confusion was around $f^{-1}[B] = \{x \in X: f(x) \in B\}$. I was thinking about these things not as sets but as functions and what I am gathering, in fact, is that if $B$ is entirely outside of the range of $f$ then $f^{-1}[B] \to \emptyset$ $\endgroup$ – Jeff Dec 24 '17 at 23:09
  • $\begingroup$ @Jeff Indeed if $B \cap f[X] = \emptyset$, then $f^{-1}[B] = \emptyset$, and more generally $f^{-1}[B] = f^{-1}[B \cap f[X]]$ is true. $\endgroup$ – Henno Brandsma Dec 25 '17 at 6:42
  • $\begingroup$ @Jeff As an illustrative example: Every constant function is continuous between any pair of spaces: if $f$ has constant value $p \in Y$, then $f^{-1}[B]$ is either $\emptyset$ (iff $p \notin B$) or $X$ (iff $p \in B$) so always open in $X$. $\endgroup$ – Henno Brandsma Dec 25 '17 at 10:00

Here's an ambiguity:

  • If $y$ is a point in the image of $f,$ then $f^{-1}(y)$ means the unique point, if any, in the domain of $f$ for which $f(x) = y.$ If there is no such point or more than one such point, then the expression $f^{-1}(y)$ is undefined.

  • But if $B$ is a subset, rather than a member, of the image of $f,$ the $f^{-1}(B)$ means the set $A= \{\, x : f(x) \in B \,\}.$ There is no subset $B$ of the image for which $f^{-1}(B)$ is undefined. That's what you're talking about when you say $f$ is continuous if and only if for every open set $U$ in the codomain, $f^{-1}(U)$ is open in the domain.

Some writers on set theory will call this $f[B]$ as opposed to $f(B).$ This is important when $B$ is both a member and a subset of the domain, as sometimes happens in set theory. In that case $f[B]$ and $f(B)$ are different things.

  • $\begingroup$ Re: the first bulletpoint, I've also frequently seen the unfortunate use of "$f^{-1}(y)$" for "$f^{-1}(\{y\})$." $\endgroup$ – Noah Schweber Dec 24 '17 at 20:48
  • $\begingroup$ My key concern was applying $f^{-1}(Y)$ to a point that wasn't in the image. Is that also undefined. And can we say things like $f^{-1}(B) = A$ when part of B is undefined or more precisely part of B is outside of the range of the original function $f$? $\endgroup$ – Jeff Dec 24 '17 at 21:03
  • $\begingroup$ @Jeff : If there are no points $x$ in the domain of $f$ for which $f(x) \in B,$ then $f^{-1}(B) = \varnothing,$ which is a well defined subset of the domain. $\endgroup$ – Michael Hardy Dec 25 '17 at 6:10

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