# Why are we looking for such a function?

Show that for any sets $A_1, A_2, B_1, B_2, A_1 \sim A_2 \wedge B_1 \sim B_2 \rightarrow A_1^{B_1} \sim A_2^{B_2}$.

Let $A$ set with $card(A)=m$.

Let $B$ set with $card(B)=n$.

We define $m^n:=card(A^B)$.

Let $f: A_1 \to A_2$, $g: B_1 \to B_2$ two bijective functions.

We are looking for a $T: A_1^{B_1} \to A_2$ that is bijective.

We are we looking for a bijective function from $A_1^{B_1}$ to $A_2$ and not from $A_1^{B_1}$ to $A_2^{B_2}$ ?

• Because professor has nothing better for you to do... – Shahar Jan 22 '15 at 2:47
• @Shahar Do you have an idea? – evinda Jan 22 '15 at 2:48
• what does $A_1\sim~A_2$ mean? – Zero Jan 22 '15 at 2:49
• $A_1 \sim A_2$ means that these two sets have the smae cardinality. – evinda Jan 22 '15 at 2:50

It can happen that $|A_1^{B_1}|\neq|A_2|$ (take $A_1=A_2=B_1=B_2=\{0,1\}$ for example), but as you said we are actually looking for $|A_1^{B_1}|=|A_2^{B_2}|$, such a bijection is provided by $\phi:A_1^{B_1}\to A_2^{B_2},\phi(h)=f^{-1}\circ h\circ g$.