I'm reading Finite Sets from Munkres' Topology where he discusses a lemma:
Lemma 6.1 Let $n$ be a positive integer. Let $A$ be a set; let $a_0$ be an element of $A.$ Then there exists a bijective correspondence $f$ of the set $A$ with $\{1 \ldots n+1\}$ if and only if there exists a bijective correspondence $g$ of the set $A-\{a_0\}$ with the set $\{1\ldots n\}\,.$
First the author assumes that $f(a_0) = n+1,$ then it's obvious that $f: A-\{a_0\}\mapsto \{1\ldots n\}\,.$
Now, if above is not the case, then he lets $f(a_0) = m$ and $f(a_1)= n+1\,.$
Then he defines a new function $h$ and shows that it is bijective.
I'm having problem in comprehending the last part of the proof.
What is the point of letting $f(a_0) = m$ and $f(a_1) = n+1$ in defining $h$ and showing that $h$ is bijective?
Also, suppose $f(a_0) = m\in S= \{1\ldots n+1\};$ this means there are $n$ elements left in $S\,.$ Since $f$ was assumed to be bijective, won't that be obvious that $f: A-\{a_0\}\mapsto \{1\ldots n+1\}- \{m\}$ is a bijective correspondence? What is the necessity of defining another new function $h\,?$
The point is that a bijection $f\colon A\to B$ is a function whose domain is the entire set $A$, its range is the entire set $B$, and the function is injective.
If $f(a_0)=n+1$, then restriction $f$ to $A\setminus\{a_0\}$ does indeed provide us with a function whose range is $\{1,\ldots,n\}$. But if $f(a_0)=1$, for example, then restricting $f$ to $A\setminus\{a_0\}$ results in a function whose range is $\{2,\ldots,n+1\}$ instead.
It is true that $\{1,\ldots,n\}$ and $\{2,\ldots,n+1\}$ have the same size and there is an easy bijection between them. But you need to prove the statement that you want to prove, and the statement is to find a bijection with $\{1,\ldots,n\}$. So you might need to compose $f$ with a permutation of $A$ or of $\{1,\ldots,n+1\}$ to ensure that $f(a_0)=n+1$.