Construction of bijection between set of functions $f : X \leftrightarrow Y$ and i want to construct bijection between $\lbrace g :     
X \rightarrow A \rbrace$ and $\lbrace g : Y \rightarrow A \rbrace$.
How to do this?
 A: Suppose $g:X\rightarrow A$.  Then $g\circ f^{-1}: Y \rightarrow A$.  So, I guess you need the mapping
$$g \mapsto g \circ f^{-1}.$$
A: HINT: The bijection $f$ matches up points of $X$ with points of $Y$: you can think of $Y$ as being in a sense just another copy of $X$ under a different name, with $f$ giving each $x\in X$ an alias $f(x)\in Y$. Now in essence you want to give each function $g:X\to A$ an alias $F(g):Y\to A$ that makes it the ‘same’ function as $g$ but using the aliases in $Y$ instead of the original points in $X$. There’s only one reasonable way to do this; can you see what it is?
A: Let $B := \{g: X \to A \}, C:= \{g: Y \to A \}.$ Define $\Phi : C \to B$ by $\Phi (g) = gf.$
$\Phi$ is injective: Let $g_1, g_2 \in C$ is such that $g_1f = g_2f.$ So $g_1f(x) = g_2f(x), \forall x \in X.$ Now every element of $X$ can be uniquely written as $f^{-1}(y)$ for some $y \in Y.$ This shows that $g_1(y) = g_2(y), \forall y \in Y.$ So $g_1 = g_2.$
$\Phi$ is surjective: Let $g \in B.$ Define $h: X \to A$ as follows: $x \mapsto g(f(x)).$
A: $\require{AMScd}$
\begin{CD}
    X @>f>> Y\\
    @V \Phi(g) V V\# @VV g V\\
    A @= A
    \end{CD}
$\Phi(g)=g\circ f$ is an isomorphism $Func(Y,A)\to Func(X,A)$ mapping $g\mapsto g\circ f$
