Proving that an injective function is bijective I am having a lot of trouble starting this proof. I would greatly appreciate any help I can get here. Thanks.
Let $n\in \mathbb{N}$. Prove that any injective function from $\{1,2,\ldots,n\}$ to $\{1,2,\ldots,n\}$ is bijective.
 A: define a new function
$$
g: \operatorname{Im}{f} \rightarrow \{1, \cdots, n\}
$$
by setting $g(y)$ as that $x$ such that $f(x)=y$ (well-defined because each $y$ is an image and $f$ is injective). note that $f\circ g$ is the identity on $\operatorname{Im}{f}$, hence g must be injective; likewise, $g\circ f$ is the identity on $\{1, \cdots, n\}$, hence g must be surjective. we have just proved that g is a bijection, i.e. a permutation of $1, \cdots, n$.
the concept of cardinality is just shorthand for 'there exists a bijection'...
A: Another hint: Prove it by induction. It’s clear for $n=1$. Otherwise if the statement holds for some $n$, take an injective map $σ \colon \{1, …, n+1\} → \{1, …, n+1\}$. Assume $σ(n+1) = n+1$ – why can you do this? What follows?
A: Hint: Let $f : [n] \to [n]$ be injective.  What is the cardinality of the image of $f$?
A: Hint: If $f:[n]\rightarrow [n]$ is injective (where $[n]= \{1,2,\dots,n\}$), all that remains to be shown is that $f$ is surjective. So, suppose it's not. How does the size of the image compare to the size of the domain, and what does this say about injectivity?
A: What if $f$ would not be bijective? Then one number would not be in the image of $f$. How can that be?
A: Suppose $f : \{1,\ldots, n\}\to\{1,\ldots, n\}$ is injective. Then $a\neq b\implies f(a)\neq f(b)$, so $\left|\,\operatorname{Im}f\,\right| = \left|\{1,\ldots, n\}\right| = n$. As $\{1,\ldots,n\}$ is the codomain, what can we say?

Since $\{1,\ldots, n\}$ is the codomain and $\left|\{1,\ldots, n\}\right| = n$, everything in the codomain must be hit.

