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.
f: A→Bf:\ A\rightarrow Bf: A→B
1. ∀a∈A:(f∘IA)(a)=f(IA(a))\forall a \in A:\quad (f\circ I_A)(a)=f\big(I_A(a)\big)∀a∈A:(f∘IA)(a)=f(IA(a))
Since a∈Aa\in Aa∈A , it follows that IA(a)=aI_A(a)=aIA(a)=a and (f∘IA)(a)=f(IA(a))=f(a)(f\circ I_A)(a)=f\big(I_A(a)\big)=f(a)(f∘IA)(a)=f(IA(a))=f(a)
So, f∘IA=ff\circ I_A=ff∘IA=f .
2. ∀a∈A:(IB∘f)(a)=IB(f(a))\forall a \in A:\quad (I_B\circ f)(a)=I_B\big(f(a)\big)∀a∈A:(IB∘f)(a)=IB(f(a))
Since f(a)∈Bf(a)\in Bf(a)∈B , it follows that IB(f(a))=f(a)I_B\big(f(a)\big)=f(a)IB(f(a))=f(a) and (IB∘f)(a)=f(a)(I_B\circ f)(a)=f(a)(IB∘f)(a)=f(a)
So, IB∘f=fI_B\circ f=fIB∘f=f .
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