Set Equivalence Less One Element

Theorem
Let $$S$$ and $$T$$ be sets such that $$S \sim T$$, i.e. they are equivalent.

Let $$a \in S, b \in T$$.

Then $$S - \left\{{a}\right\} \sim T - \left\{{b}\right\}$$

Proof
As $$S \sim T$$, there exists a bijection $$f: S \to T$$.

We define the mapping $$g: \left({S - \left\{{a}\right\}}\right) \to \left({T - \left\{{b}\right\}}\right)$$ as follows:

$$ \forall x \in S - \left\{{a}\right\}: g \left({x}\right) = \begin{cases} f \left({x}\right): x \ne b \\ f \left({a}\right): x = b \end{cases} $$

It is easily confirmed that this is a bijection.