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Archangel Macsika

Let f : A → B be a function. 1. Show that for the identity function iA on A we have f ◦ iA = f. 2. Show that for the identity function iB on B we have iB ◦ f = f.

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f:\ A\rightarrow B

1. \forall a \in A:\quad (f\circ I_A)(a)=f\big(I_A(a)\big)

Since a\in A , it follows that I_A(a)=a and (f\circ I_A)(a)=f\big(I_A(a)\big)=f(a)

So, f\circ I_A=f .


2. \forall a \in A:\quad (I_B\circ f)(a)=I_B\big(f(a)\big)

Since f(a)\in B , it follows that I_B\big(f(a)\big)=f(a) and (I_B\circ f)(a)=f(a)

So, I_B\circ f=f .


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