Composite function by definition

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April 27th 2010, 11:33 AM
novice

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:
$ (g \circ f)(x)=g(f(x))=g(y)=z $ 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 $y \in Y$ .

$ 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

$ id_Y \circ g =? $

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?
April 27th 2010, 12:03 PM
Plato

Quote:

Originally Posted by novice

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?
April 27th 2010, 12:14 PM
novice

That is so clever.(Clapping)