Show that $(g \circ f)^{-1}(C) = g^{-1}(f^{-1}(C)).$

Question

Suppose that $f:S \rightarrow T$ and $g:T \rightarrow U$ are two functions and that $C \subseteq U.$ Show that $(g \circ f)^{-1}(C) = g^{-1}(f^{-1}(C)).$

Is the statement of this question is correct? or the correctness depends on whether the $(-1)$ represents the inverse function or the preimage (inverse image)?

It's not correct. It should be $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ for the preimage. — Qi Zhu, Jan 06 '20 at 11:17
Note that $C$ is not even in $T$, so $f^{-1}(C)$ is not well-defined. — Qi Zhu, Jan 06 '20 at 11:17
Related: Prove that if $Z\subseteq Y$, then $(g\circ f)^{-1}(Z)=f^{-1}(g^{-1}(Z))$. — Martin Sleziak, Jan 06 '20 at 12:15

José Carlos Santos · Accepted Answer · 2020-01-06T11:29:36.483

7

The statement doesn't make sense, since $C\subset U$ and $f$ is a map from $S$ into $T$. Therefore, $f^{-1}(C)$ is undefined.

However, it is true that$$(g\circ f)^{-1}(C)=f^{-1}\bigl(g^{-1}(C)\bigr).$$In fact\begin{align}x\in(g\circ f)^{-1}(C)&\iff g\bigl(f(x)\bigr)\in C\\&\iff f(x)\in g^{-1}(C)\\&\iff x\in f^{-1}\bigl(g^{-1}(C)\bigr).\end{align}

edited Jan 06 '20 at 11:29

answered Jan 06 '20 at 11:21

José Carlos Santos

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Do we need an extra assumption that $f,g$ are 1-1 and Onto as I saw this extra assumption found in Schaum Series of "general topology" on pg.25 – Intuition Jan 08 '20 at 05:41
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No, we don't. Note that, here, $f^{-1}$ has nothing to do with inverse function. In fact, $f^{-1}(C)$ is the inverse image of $C$. – José Carlos Santos Jan 08 '20 at 07:14

Show that $(g \circ f)^{-1}(C) = g^{-1}(f^{-1}(C)).$

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