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, $I_B\circ f=f$ .