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Let $\mathcal{F}(A,\,B)=\{f \text{ function}\,:\,f:A\to B \}$, so I have to determine a bijection between $\mathcal{F}(A,\,B)$ and $\mathcal{F}(A-\{a\},\,B)$, with $a\in A$.

I'm having a little trouble with this one, every help will be really appreciated.

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    In general there doesn't need to be a bijection. If $A=\{0,1\}$ and $B=\{0,1\}$, then $|\mathcal{F}(A,B)| = 4$ while $|\mathcal{F}(A\setminus\{1\},B)| = 2$. More generally, if $|A|=n$ and $|B|=m$ then $|\mathcal{F}(A,B)| = m^n$, so $|\mathcal{F}(A\setminus\{a\},B)| = m^{n-1}$. Thus, you won't get a bijection so long as $m\neq 0$ and $n,m$ are finite. If $A$ is infinite, then you will get a bijection. – 2017-02-20
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    Thanks! I really made this counterexample but I was explicitly making all the functions and I get confused with the functions of functions. – 2017-02-20

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