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Math Help - Composite function by definition

  1. #1
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    Composite function by definition

    I found this expression on Wikipedia:

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

    where id_X and 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:
     <br />
(g \circ f)(x)=g(f(x))=g(y)=z<br />
for f:X \rightarrow Y, g: Y \rightarrow Z, x \in X, y \in Y, and z \in Z.

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

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

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

     <br />
id_Y \circ g =?<br />

    I can't find any information anywhere on the web proving 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 f: X \rightarrow Y, then f \circ id_X = id_Y \circ f,
    where id_X and id_Y are identity functions.
    Use i_X~\&~i_Y for the identity functions.
    The by definition if (p,q)\in f\circ i_X then  \left( {\exists r} \right)\left[ {(p,r) \in i_X  \wedge (r,q) \in f} \right]
    But  (p,r) \in i_X implies p=r. So  (p,q) \in f.

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