How is the inverse of the inclusion map defined? 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 $X$.
 A: 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.
A: 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.
