Equinumerosity

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Question: If $A_1 =_c A_2$ and $B_1 =_c B_2$ prove that $(A_1 \to B_1 ) =_c (A_2 \to B_2 )$. Or in other words If the cardinality of $A_1$ and $B_1$ are equal to that of $A_2$ and $B_2$ respectively, then the cardinality of the set of functions from $A_1 \to B_1$ is equal to that of the set of functions from $A_2 \to B_2$ (from Notes on Set Theory, Yiannis Moschovakis, Chapter 2, Exercise 2.33)

Proof: Let $\sigma : A_1 \to A_2$ and $\phi : B_1 \to B_2$ be two bijections. And $f: A_1 \to B_1$ be any map. Define $g = \phi \cdot f \cdot \sigma^{-1}$. We claim that this map $T:(f \to g)$ is a bijection.

Is $T$ well defined?

Verify that $T(f):=\phi \cdot f \cdot \sigma^{-1}$ is indeed a map from $(A_2 \to B_2)$

Is $T$ onto?

Let $g_0$ be any arbitary map from $(A_2 \to B_2)$. Then $T(\phi^{-1} \cdot g_0 \cdot \sigma) = \phi \cdot \phi^{-1} \cdot g_0 \cdot \sigma \cdot \sigma^{-1} = g_0$

Is $T$ one-one?

Let $T(f_1) = T(f_2)$

$\implies \quad \phi \cdot f_1 \cdot \sigma^{-1} = \phi \cdot f_2 \cdot \sigma^{-1} \ \implies \quad \phi^{-1} \phi \cdot f_1 \cdot \sigma^{-1} \sigma \ = \ \phi^{-1} \phi \cdot f_2 \cdot \sigma^{-1} \sigma \ \implies \quad f_1 = f_2 \$

Note that since $\phi$ and $\sigma$ are bijections, their inverses are well defined and function property of equal input gives equal output.

Hence $T$ is a bijection from $(A_1 \to B_1 ) \to (A_2 \to B_2 )$ and therefore $(A_1 \to B_1 ) =_c (A_2 \to B_2 )$