# mapping question

• October 10th 2008, 09:53 PM
lllll
mapping question
let $f$ be a function from $X$ into $Y$. Let g be a function from $Y$ into $Z$. Let $B \subset Z$. Show that $(g \circ f)^{-1}(B) = f^{-1}(g^{-1}(B))$

Assuming that both function are injective, when applying the inverse you would get something along the lines of:

$Z \mapsto Y \mapsto X$

$f^{-1}(g^{-1}(B_1)) = f^{-1}(g^{-1}(B_2)) \rightarrow g^{-1}(B_1) = g^{-1}(B_2) \rightarrow B_1=B_2.$

at least for the right hand side, showing the left is somewhat more difficult.
• October 11th 2008, 06:16 AM
Moo
$\begin{gathered} \text{Hello,} \hfill \\ \\
\text{I don't see why you assume it is injective...} \hfill \end{gathered}$

$\begin{gathered} \text{Remember that if a function } f ~:~ X \to Y~, \hfill \\
\text{Then if }\mathcal{Y} \subset Y ~,~ f^{-1}(\mathcal{Y})=\{x \in X ~:~ f(x) \in \mathcal{Y}\} \hfill \end{gathered}$

$\begin{gathered} X \stackrel{f}{\longrightarrow} Y \stackrel{g}{\longrightarrow} Z \quad \quad \quad \quad \hfill Z \stackrel{g^{-1}}{\longrightarrow} Y \stackrel{f^{-1}}{\longrightarrow} X \\
X \stackrel{f \circ g}{\longrightarrow} Z \hfill Z \stackrel{(f \circ g)^{-1}}{\longrightarrow} X \\
\end{gathered}$

________________________________________________

$(g \circ f)^{-1}(B)=\big\{x \in X ~:~ (g \circ f)(x) \in B \big\}$

$\begin{gathered} \mathcal{B}=g^{-1}(B)=\big\{y \in Y ~:~ g(y) \in B\big\} \hfill \\
\Rightarrow f^{-1}(g^{-1}(B))=f^{-1}(\mathcal{B})=\big\{x \in X ~:~ f(x) \in \mathcal{B}\big\} \hfill \\
=\big\{x \in X ~:~ g(f(x)) \in B \big\} \text{ since } f(x) \in \mathcal{B} \hfill \\ \end{gathered}$

$\Longrightarrow f^{-1}(g^{-1}(B))=(g \circ f)^{-1}(B)$

Uuuum does it look clear ? (or correct ? :p)
• October 11th 2008, 07:02 AM
Plato
Let’s do this for relations in general. Every function is a relation.
$\begin{gathered} \left( {p,q} \right) \in f^{ - 1} \circ g^{ - 1} \; \Rightarrow \;\left( {\exists r} \right)\left[ {\left( {p,r} \right) \in g^{ - 1} \wedge \left( {r,q} \right) \in f^{ - 1} } \right] \hfill \\ \left( {q,r} \right) \in f \wedge (r,p) \in g\; \Rightarrow \;(q,p) \in g \circ f \hfill \\ (q,p) \in g \circ f\; \Rightarrow \;(p,q) \in (g \circ f)^{ - 1} \hfill \\
\end{gathered}$
.
That is one direction. The other is about the same in a little different order.