Bijection between a set of injections and a union of bijection Consider $A$ a set with $n$ elements and $k\leq n$.
Show that
$$
\phi:\left\{  f:\left\{  1,2,\ldots,k\right\}  \rightarrow A\mid f\text{
injective}\right\}  \rightarrow
{\bigsqcup\limits_{B\subset A,card(B)=k}}
\left\{  g:\left\{  1,2,\ldots,k\right\}  \rightarrow B\mid g\text{
bijective}\right\}
$$
by constructing an explicit bijection.
By generalizing the construction above, show that
$$
\Phi:\left\{  f:A\rightarrow C\mid f\text{ injective}\right\}  \rightarrow
{\bigsqcup\limits_{B\subset C.card(C)=n}}
\left\{  g:B\rightarrow C\mid g\text{ bijective}\right\}  .
$$
I attempted to start from the observation that every injective function $f$
has an image $\operatorname{Im}f$ as subset of $A$ with exactly $k$ elements.
Reciprocally every subset of $A$ with $k$ elements is the image of some
injection. But there are $k!$ injections that maps a fixed set $C$ into a
subset of $A$ with $k$ elements. Now, every
$$
f:\left\{  1,\ldots,k\right\}  \rightarrow A
$$
gives us a bijection $g:\left\{ 1,\ldots,k\right\}  \rightarrow B,B\subset A$
and $card(B)=k$. From this point, I don't know how to move forward. How can I
find an explicit bijection?
 A: Hint
Denote $E=\left\{  f:\left\{  1,2,\ldots,k\right\}  \rightarrow A\mid f\text{
injective}\right\}$ and $F={\bigsqcup\limits_{B\subset A,card(B)=k}}
\left\{  g:\left\{  1,2,\ldots,k\right\}  \rightarrow B\mid g\text{
bijective}\right\}$.
Let
$$\begin{array}{l|rcl}
\Phi : & E & \longrightarrow & F \\
    & f & \longmapsto & \begin{array}{l|rcl}
\Phi(f) : & \left\{  1,2,\ldots,k\right\} & \longrightarrow & f[\left\{  1,2,\ldots,k\right\}] \\
    & x & \longmapsto & f(x)  \end{array}  \end{array}
$$
$\Phi$ is a formal definition of the bijection you're looking for.
A: $\require{HTML}$You basically answered your own question already.
Given sets $|X|$ and $|Y|$ with $|X|\le |Y|$ we have a bijection
\begin{align}
\{\text{injections $X\to Y$}\} &\longleftrightarrow \bigsqcup_{\substack{Z\subset Y,\\|Z|=|X|}} \{\text{bijections $X\to Z$}\}, \\
(f\colon X\to Y) &\longmapsto (X\to \operatorname{Im}(f), x\mapsto f(x)),\\
(X\to Y, x\mapsto g(x))&\,\style{display: inline-block; transform: scaleX(-1)}{\longmapsto} (g\colon X\to Z).
\end{align}
