$S \subseteq T \Rightarrow |A^S| \leq |A^T|$ only when $A \neq \emptyset$ Exercise 1.7 from Chapter 4 of Hrbacek & Jech's Introduction to Set Theory (3rd Ed.) asks to prove:
If $S \subseteq T$, then $|A^S| \leq |A^T|$.
I think this is true only when $A \neq \emptyset$. Am I right?

Remark: This question was posted with the purpose of being answered by myself. You can see my own answer below, but that is not completely right, as was pointed out by M. Logic.
 A: Although there is a typo in the exercise, you are not completely right. Since $S\subseteq T$, there are three subcases for the case $A=\varnothing$.
Subcase 1.1 $S=T=\varnothing$. The proposition is right since $A^S=A^T=\{\varnothing\}=\{0\}=1$.
Subcase 1.2 $S=\varnothing$ and $T\neq\varnothing$. The proposition is false as you shown.
Subcase 1.3 $S\neq\varnothing$ and $T\neq\varnothing$. The proposition is right since $A^S=A^T=\varnothing=0$.
And for the case $A\neq\varnothing$, there are also three subcases.
Subcase 2.1 $S=T=\varnothing$. The proposition is right since $A^S=A^T=\{\varnothing\}=\{0\}=1$.
Subcase 2.2 $S=\varnothing$ and $T\neq\varnothing$. Pick $a\in A$, then the constant function $\mathbf{a}(x)$ defined by $\mathbf{a}(x)=a$ for all $x\in T$ is in $A^T$. And since $A^S=\{\varnothing\}=\{0\}=1$ we have $|A^S|\leq|A^T|$.
Subcase 2.3 $S\neq\varnothing$ and $T\neq\varnothing$. The proposition is right as you shown.
A: Yes, you is right. Consider $A = \emptyset$, $S = \emptyset$, $T = \{\emptyset\}$. Then, $S \subseteq T$, but $A^S = \emptyset^\emptyset = \{\emptyset\}$, whilst $A^T = \emptyset^{\{\emptyset\}} = \emptyset$, so that there is no function on $A^S$ into $A^T$. In particular, there is no injection on $A^S$ into $A^T$. That is, $|A^S| \not\leq |A^T|$.
On the other hand, if $A \neq \emptyset$, then there is an $a \in A$ for which the map $$\begin{aligned}\theta: A^S &\to A^T\\f &\mapsto \theta_f\end{aligned}$$
given by $$\theta_f(x) = \begin{cases} f(x) &\mbox{ if $x \in S$,}\\ a &\mbox{ otherwise} \end{cases}$$ is an injection on $A^S$ into $A^T$.
Conclusion: there is a typo in that exercise. The author missed the hypotesis that $A \neq \emptyset$.
