Results 1 to 3 of 3

Math Help - mapping question

  1. #1
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1

    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.
    Last edited by lllll; October 10th 2008 at 11:10 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    \begin{gathered} \text{Hello,} \hfill \\ \\<br />
\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 \\<br />
\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 \\<br />
X \stackrel{f \circ g}{\longrightarrow} Z \hfill Z \stackrel{(f \circ g)^{-1}}{\longrightarrow} X \\<br />
\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 \\<br />
\Rightarrow f^{-1}(g^{-1}(B))=f^{-1}(\mathcal{B})=\big\{x \in X ~:~ f(x) \in \mathcal{B}\big\} \hfill \\<br />
=\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 ? )
    Last edited by Moo; October 11th 2008 at 07:33 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1
    Letís 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 \\ <br />
\end{gathered} .
    That is one direction. The other is about the same in a little different order.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mapping question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: June 14th 2011, 07:07 AM
  2. Complex open mapping & conformal mapping problems.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 22nd 2011, 08:26 AM
  3. simple question on mapping from R^n -> R^m
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: June 1st 2010, 11:13 PM
  4. K-mapping question
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 1st 2009, 06:37 PM
  5. another mapping question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 13th 2008, 04:04 PM

Search Tags


/mathhelpforum @mathhelpforum