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This question is related to one I asked here.

I am trying, overall, to show that $|A - \{x\}| =n-1$ using a bijection given that $A$ is a finite set with $|A|\geq 1$. For the case when $n > 1$, suppose that, since $|A|=n$, $\exists$ a bijection $f: A \to \mathbb{N}_{n}$.

With some help, I came up with the mapping $f^{-1}\circ \tau \circ I: \mathbb{N}_{n-1} \to A - \{x\}$.

Here, $I: \mathbb{N}_{n-1} \to \mathbb{N}_{n}$ where for each $j \in \mathbb{N}_{n-1}$, $I(j) = j$. It is not difficult to show that $I$ is injective, but it is not surjective, because $n$ has no preimage.

$\tau: \mathbb{N}_{n} \to \mathbb{N}_{n}$, where $\forall k \in \mathbb{N}_{n}$, $\tau(k) = \begin{cases} n & \text{if} \, k=f(x) \\ f(x) & \text{if} \, k = n \\ k & \text{otherwise} \end{cases}$

$\tau$ is bijective, whether we restrict it to $\mathbb{N}_{n} - \{f(x)\}$ or not.

Now, I need to show that the composition $f^{-1} \circ \tau \circ I$ is a well-defined bijection in order to complete my cardinality proof.

Since $I$, $\tau$, and $f^{-1}$ (since $f$ is, its inverse is) are all injective, their composition is injective.

Showing surjectivity and that the map is well-defined are not so easy for me, however. To show that $f^{-1} \circ \tau \circ I$ is surjective, I know that I need to show that every $y \in A - \{ x \}$ has a preimage in $\mathbb{N}_{n-1}$, but I'm not specifically sure how to do that. Could someone please help me with this part?

Also, to show that it's well-defined, I need to show that two different $i$'s in $\mathbb{N}_{n-1}$ do not map to the same $y \in A - \{ x \}$. Suppose they did. I.e., suppose that $i \neq j$ and $f^{-1}(\tau(I(i))) = f^{-1}(\tau(I(j)))$. Since the map is injective, it seems like this shouldn't work, but I can't use that to help me prove well-definedness, so I could use some help with this part as well.

Thank you.

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    Another part of well-definedness is verifying that your function actually maps into $A-\{x\}$: this doesn't come free by composition because $f^{-1}$ maps into $A$. For the first bolded question, suppose $y \in A-\{x\}$. What can you say about the value of $f(y)$? This tells you something about $\tau^{-1}(f(y))$. From there you can deduce that the latter has a preimage under $I$, which is what you need for surjectivity. – 2017-02-10

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I will fornish to you a (maybe) different proof.

I think for you $\mathbb{N}_n:=\{1,2,...,n\}$. For me it is.

Suppose that $|A|=n$. So there is a bijection $f:\mathbb{N}_n\to A$. Since $x\in A$ and $f$ is surjective, there is $i\in\mathbb{N}_n$ such that $f(i)=x$. Define $\tau: \mathbb{N}_{n} \to \mathbb{N}_{n}$, $\forall k \in \mathbb{N}_{n}$, by

$$\tau(k) = \begin{cases} k & \text{if} ~~ 1\leq k < i \\ k-1 & \text{if} ~~ i < k \leq n \\ n & \text{if} ~~ k = i, \end{cases}$$

So $\tau$ "translade" the point $i$ (such that $f(i)=x$) to the "end of the set" $\mathbb{N}_{n}$. It is an easy exercise to show that $\tau$ is a bijection. So the function $f\circ \tau$ is still a bijection, as a composition of two bijections. Furthermore, $(f\circ \tau)(n)=x$. So if you define $\bar{f}=f\circ \tau|_{\mathbb{N}_{n-1}}$ (the restriction of $f\circ \tau$ to the set $\mathbb{N}_{n-1}$) you will get your desired bijection.

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    thank you! This is much better than what I had originally! – 2017-02-12