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Thread: Composite function by definition

  1. #1
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    Composite function by definition

    I found this expression on Wikipedia:

    If $\displaystyle f: X \rightarrow Y$, then $\displaystyle f \circ id_X = id_Y \circ f$,

    where $\displaystyle id_X$ and $\displaystyle id_Y$ are identity functions.

    I wanted to see whether it's true. So I rewrite the expression using the definition of composite function.

    Definition:
    $\displaystyle
    (g \circ f)(x)=g(f(x))=g(y)=z
    $ for $\displaystyle f:X \rightarrow Y$, $\displaystyle g: Y \rightarrow Z$, $\displaystyle x \in X, y \in Y$, and $\displaystyle z \in Z$.

    I replaced $\displaystyle f$ with $\displaystyle g$ in $\displaystyle f \circ id_X$ and replaced $\displaystyle id_X$ with $\displaystyle f(x)=x$, where $\displaystyle x \in X$. So I wrote the expression for the LHS

    Let $\displaystyle f:X\rightarrow X$ and $\displaystyle g:x\rightarrow Y$, where $\displaystyle x\in X $and $\displaystyle y \in Y$.

    $\displaystyle
    g\circ id = (g\circ f)(x)=g(f(x))=g(x)=y= g
    $, but now my difficulty being that I don't know how to express the RHS

    $\displaystyle
    id_Y \circ g =?
    $

    I can't find any information anywhere on the web proving $\displaystyle f \circ id_X = id_Y \circ f$.

    Could someone please show me how to write the expression for the RHS?
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  2. #2
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    Quote Originally Posted by novice View Post
    If $\displaystyle f: X \rightarrow Y$, then $\displaystyle f \circ id_X = id_Y \circ f$,
    where $\displaystyle id_X$ and $\displaystyle id_Y$ are identity functions.
    Use $\displaystyle i_X~\&~i_Y$ for the identity functions.
    The by definition if $\displaystyle (p,q)\in f\circ i_X$ then $\displaystyle \left( {\exists r} \right)\left[ {(p,r) \in i_X \wedge (r,q) \in f} \right]$
    But $\displaystyle (p,r) \in i_X $ implies $\displaystyle p=r$. So $\displaystyle (p,q) \in f$.

    If $\displaystyle (p,q)\in i_Y\circ f$ then $\displaystyle \left( {\exists s} \right)\left[ {(p,s) \in f \wedge (s,q) \in i_Y} \right]$.
    $\displaystyle (s,q) \in i_Y $ implies that $\displaystyle s=q$, so $\displaystyle (p,q)\in f$.

    Does that help?
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  3. #3
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    That is so clever.
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