Why $\boldsymbol{n} \approx \boldsymbol{n}^{+}$ contradicts the pigeon-hole theorem? Can anyone draw a contradiction using the pigeon-hole theorem from $\boldsymbol{n} \approx \boldsymbol{n}^{+}$?
Pigeon-hole theorem is the following:

For all $\boldsymbol{n} \in \mathbb{N}$, $f: \boldsymbol{n} \to \boldsymbol{n}$ is onto if it is one-one.

$\boldsymbol{n}^{+} = \boldsymbol{n} \cup \left\{ \boldsymbol{n} \right\}$ is the successor of $\boldsymbol{n} \in \mathbb{N}$.
$\boldsymbol{n} \approx \boldsymbol{n}^{+}$ means there is a bijection between $\boldsymbol{n}$ and $\boldsymbol{n}^{+}$.
The formula $\boldsymbol{n} \approx \boldsymbol{n}^{+}$ may seem naively false, as $\boldsymbol{n}^{+}$ is of one more element than $\boldsymbol{n}$. But ironically, I find it difficult to prove that the formula is false.

I have used the approach provided in this post to prove a similar, but more general theorem:

If $\boldsymbol{m} < \boldsymbol{n}$, then $\boldsymbol{m} \prec \boldsymbol{n}$.

Assume $\boldsymbol{m} < \boldsymbol{n}$. Then $\boldsymbol{m} \subsetneq \boldsymbol{n}$. Define the identity map $f: \boldsymbol{m} \to \boldsymbol{n}$ as
\begin{equation*}
\left(\forall \boldsymbol{s} \in \boldsymbol{m}\right) \left(f\left(\boldsymbol{s}\right) = \boldsymbol{s}\right).
\end{equation*}
It is immediately clear that $f$ is one-one. Thus, $\boldsymbol{m} \preceq \boldsymbol{n}$.
Now assume that $\boldsymbol{m} \approx \boldsymbol{n}$. Then there exists a bijection $g: \boldsymbol{n} \to \boldsymbol{m}$. The restriction of $g$ to $\boldsymbol{m}$, $g\vert_{\boldsymbol{m}}: \boldsymbol{m} \to \boldsymbol{m}$ has to be one-one, thus it also has to be onto, according to pigeon-hole theorem. As $\boldsymbol{m} \subsetneq \boldsymbol{n}$, there exists $\boldsymbol{p} \in \mathbb{N}$, $\boldsymbol{p} \in \boldsymbol{n}$ and $\boldsymbol{p} \not\in \boldsymbol{m}$. As $g$ is onto, $g\left(\boldsymbol{p}\right) \in \boldsymbol{m}$. As $g\vert_{\boldsymbol{m}}$ is onto and $\boldsymbol{p} \not\in \boldsymbol{m}$, there has to exist $\boldsymbol{q} \in \boldsymbol{m}$ such that $\boldsymbol{q} \neq \boldsymbol{p}$ and $g\vert_{\boldsymbol{m}}\left(\boldsymbol{q}\right) = g\left(\boldsymbol{p}\right)$. Using the fact that $g\vert_{\boldsymbol{m}}\left(\boldsymbol{q}\right) = g\left(\boldsymbol{q}\right)$, we have $\boldsymbol{p} \neq \boldsymbol{q}$ and $g\left(\boldsymbol{p}\right) = g\left(\boldsymbol{q}\right)$, which contradicts the assumption that $g$ is one-one. Thus, there is no bijection between $\boldsymbol{m}$ and $\boldsymbol{n}$, and $\boldsymbol{m} \not\approx \boldsymbol{n}$. Combining $\boldsymbol{m} \preceq \boldsymbol{n}$ and $\boldsymbol{m} \not\approx \boldsymbol{n}$, we may conclude that $\boldsymbol{m} \prec \boldsymbol{n}$.
 A: Suppose $f:{\bf n^+}\rightarrow {\bf n}$ is a bijection. Think about the restriction $g$ of $f$ to ${\bf n}$. Since $f$ was a bijection, $g$ must be an injection: any restriction of an injection is again an injection. But by the pigeonhole principle, this means that $g$ is onto. Now, do you see why this contradicts the injectivity of $f$?

 Since $g$ is onto and has the same codomain as $f$, we must have $f({\bf n})\in ran(g)$, which is to say $f({\bf n})=g({\bf k})$ for some ${\bf k}\in dom(g)$. But the domain of $g$ is exactly ${\bf n}$ and so ${\bf k}\not={\bf n}$. Since $g$ is a restriction of $f$ we have $f({\bf k})=g({\bf k})$, and putting all this together we get $$f({\bf n})=f({\bf k})\quad\mbox{but}\quad {\bf n}\not={\bf k}.$$


Note that this is somehow "dual" to Yuz's argument in the comments, which replaces $f$ by its "codomain-only expansion" $$\overline{f}: {\bf n^+}\rightarrow{\bf n^+}: a\mapsto f(a).$$ Ultimately these are using the same idea, but they deploy it differently.
