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I think this is a super silly question, but I just can't figure out why is that given any function $f: X \to Y$, where $(X, \mathcal{T})$ is an arbitrary topological space, and $(Y, \mathcal{T}_{trivial})$ where $ \mathcal{T}_{trivial} = \{\varnothing, Y\}$

Okay, so $f^{-1}(\varnothing) = \varnothing \in \mathcal{T}$, but how do we know that $f^{-1}(Y) \in \mathcal{T}?$ Since the preimage of the codomain is not necessarily the domain When is the preimage of codomain not equal to domain?

Why couldn't there be a case where $f^{-1}(Y) = U \subset X$, but $U \notin \mathcal{T}?$


Edit: So is it always the case that $f^{-1}(Y) = X$, given $f: X \to Y$?

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    $\begingroup$ the preimage of the codomain is the domain; everything in $X$ gets maps to $Y$. $\endgroup$ – yoyo Jun 3 '16 at 21:55
  • $\begingroup$ All the points of the domain are mapped to the codomain so the preimage of the codomain contains the domain. $\endgroup$ – Masacroso Jun 3 '16 at 22:06
  • $\begingroup$ Buf of course $f^{-1}(Y) = X$, for every function $f\colon X\to Y$. For every $x\in X$, $f(x)\in Y$, so $x \in f^{-1}(Y)$. Your title is a little confusing. There are many "trivial" (indiscrete) spaces. I thought by "the" trivial space that perhaps you meant a/the one-point space. $\endgroup$ – BrianO Jun 4 '16 at 0:00
  • $\begingroup$ @BrianO Sorry, calling it the indiscrete topology from now on $\endgroup$ – Carlos - the Mongoose - Danger Jun 4 '16 at 0:13
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By definition, for $f\colon X\to Y$ and $B\subseteq Y$, $$ f^{-1}(B)=\{x\in X:f(x)\in B\} $$ In particular, for $B=Y$, $$ f^{-1}(Y)=\{x\in X:f(x)\in Y\} $$ and therefore $f^{-1}(Y)=X$.

The question you link has a very misleading title.

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    $\begingroup$ @Lookbehindyou No need to be sorry! 😄 $\endgroup$ – egreg Jun 3 '16 at 22:17

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