- Suppose $A \precsim B$ and $C \precsim D$.
(a) Prove that if $A \neq \emptyset$ then $ {^A C } \precsim {^B D}$.
(b) Is the assumption that $A \neq \emptyset$ needed in part (a)?
Notations:
${^A C}$ is the set of all functions from $A$ to $C$.
${^B D}$ is the set of all functions from $B$ to $D$.
$A \precsim B$ means $\exists$ injection $f:A \rightarrow B$.
$C \precsim D$ means $\exists$ injection $g:C \rightarrow D$.
The answer for part (a) given in the book makes use of $A \neq \emptyset$ but gives no explanation for why $A \neq \emptyset$ is needed in the first place. So, I won't include them here.
Here is my failed attempt to see why $A \neq \emptyset$ is needed.
Looking at part (a) question, I think one possible solution is to find an example where $A = \emptyset$ and $ {^A C } \precsim {^B D}$ is false.
Suppose $A = \emptyset$.
Then the only possible $f: A=\emptyset \rightarrow B$ is $f = \emptyset$.
Thus, ${^A C} = \{ \emptyset \}$ regardless of what $C$ is.
I tried several $B$ and $D$ but found that all the $h: {^A C} \rightarrow {^B D}$ are injective (which means I can't find a case where ${^A C } \precsim {^B D}$ is false).
As an example, I tried to
let $B = \{ d \}$, $D = \{ e,f\}$, then
${^B D} = \{ \{ (d,e) \}, \{ (d,f) \} \}$
The possible $h: {^A C} \rightarrow {^B D}$ are
$h_1 = \{ (\emptyset, \{ (d,e) \} ) \}$
$h_2 = \{ (\emptyset, \{ (d,f) \} ) \}$
Both of which are injective.
As you said, when $A=\varnothing$, there is exactly one function $f=\varnothing\in {^A}C$.
In particular if $A=C=\varnothing$ this is the case, since the empty function is a function $\varnothing\to\varnothing$. However, if we let $B$ be nonempty and $D=\varnothing$, then clearly $A\precsim B$ and $C\precsim D$, but there exists no function $B\to\varnothing$.
So we see that in this case ${^A}C=\{\varnothing\}$ while ${^B}D=\varnothing$. Hence ${^A}C\precsim {^B}D$ is false.