$I_n \sim I_m $ iff $ n=m $ by induction over n This question is related to
Prove that if $A \sim I_n$ and $A \sim I_m$ then $n=m$
in which basicaly the same was proof in another way.Now my aim is to understand the following proof by mathematical induction
I have the following proof in my lecture notes:
Clearly if $n=m$, the identity mapping over $I_n$ is bijective and then $I_n \sim I_m $
Viceversa, let $I_n \sim I_m $ and let's uppose $n\leq m$. If $n=1$ and $\varphi:I_1 \rightarrow I_m$ is bijective, then it is also surjective: therefore m=1.
Let's supposse the thesis is true for a fixed $n \geq 1$ and let's verify it for $n+1$.
Let $\varphi:I_{n+1} \rightarrow I_m$ be bijective and let $u=\varphi(n+1) \in I_m$. The mapping:
$$\tau: I_m \rightarrow I_m $$
$$k \rightarrow\begin{cases} 
k, &\text{if }k \neq u,m\\
u, &\text{if } k=m\\
m, &\text{if } k=u 
\end{cases}$$
is bijective, and so is $\tau \circ\varphi:I_{n+1} \rightarrow I_m$
Because  $\tau \circ\varphi(n+1)=\tau(u)=m,$ it follows that $\varphi_{\big|I_n}$is a bijection from $I_n$ to $I_{m-1}$, so by inductive hypothesis $n=m-1$ , that is $n+1=m$. The thesis is then proved by induction
I have two questions about this proof

*

*Is there an error in the last part?: Shouldn't it be ($\tau \circ\varphi)_{\big|I_n}$ instead of $\varphi_{\big|I_n}$?


*Why do I need to define $\tau$ and $\tau \circ \varphi$ for? Can't I just define :
$\theta: I_n \rightarrow I_{m}\setminus{\varphi(n+1)}$  which is bijective since I am only taking  $(n+1,\varphi(n+1)) $ out of $\varphi$
and then apply the inductive hypothesis to it, so that
$n=m-1$ , that is $n+1=m$.
 A: You have a fairly significant typo: just under the displayed line $\tau:I_m\to I_m$ you should have $\tau(k)=$ stuff, not $k=$ stuff.
Yes, the last part should say that $(\tau\circ\varphi)\upharpoonright I_n$ is a bijection from $I_n$ to $I_{m-1}$.
In order to use the induction hypothesis, you have to have a bijection from  $I_n$ to $I_{m-1}$. If you simply remove the ordered pair $\langle n+1,\varphi(n+1)\rangle$ from $\varphi$, you have a bijection from $I_n$ to $I_m\setminus\{\varphi(n+1)\}$; unless you’re very lucky, and it just by accident happens that $\varphi(n+1)=m$, $I_m\setminus\{\varphi(n+1)\}$ is not $I_{m-1}$, so $I_n$, $\varphi\upharpoonright I_n$, and $I_m\setminus\{\varphi(n+1)\}$ do not meet the conditions required to apply the induction hypothesis.
Composing $\varphi$ with $\tau$ is just a simple way to get a bijection from $I_{n+1}$ to $I_m$ whose restriction to $I_n$ really does map to $I_m$, so that we do have the conditions required to apply the induction hypothesis.
