Assume that $X, Y$ are compact metric spaces, and that there is a map
$$ \mu : X \to \Delta (X \times Y)$$
such that $\mu$ is continuous, where $\Delta (\Omega)$ denotes the set of probability measures over a generic $\Omega$ . Endow $X \times Y$ with the product topology, and $ \Delta (X \times Y)$ with the topology of weak convergence.
The continuity of $\mu$ tells us that when we have an open (resp. closed) subset $G$ of $ \Delta (X \times Y)$, we are ensured that the preimage $\mu^{-1} (G)$ is open (resp. closed). However, I have a problem with the following situation.
Let $X$ be finite, hence compact. Let $G = \{ \delta_{(x,y)} \}$, where $\delta$ denotes the Dirac measure for some elements $x \in X$ and $y \in Y$. Thus $\mu^{-1} (\{ \delta_{(x,y)} \} )$ maps to some element $ x \in X$. But now, how can $\mu$ really be continuous in this case?
Is it continuous because we are implicitly endowing the finite $X$ with the discrete topology?
Any feedback or answer is most welcome.
Thank you for your time.
PS: To the moderators, this questions looks fairly close to this previous one. However, they are different in spirit, because that question is not well written (too many questions into one). Hence, I decided to "unzip" it, starting from this one. Regarding this issue, I think it would be wise to close that linked question (I don't know how to do it).
The constant map between two space is always continuous, in ANY topology you consider.
Suppose $\mu$ is constant, that is to say $\mu(x_1)=\mu(x_2)$ for any $x_1,x_2\in X$ and suppose $\mu(x)$ is not a dirac delta.
Then $\mu^{-1}(\delta_{(x,y)})$ is just empty, so it is not an element of $X$.
This is not a particular case because if $X$ is finite, say with $n$ elements, then the image of $\mu$ consists of just $n$ measures, and the space of probability measures on $X\times Y$ contains infinitely many elements provided $n\geq 2$ and $Y\neq\emptyset$.
Finally, if you endow $X$ with the discrete topology, then ANY map from $X$ to ANY topological space is continuous. On the other opposite, if $X$ has the trivial topology (the open sets are $X$ and $\emptyset$) then a function from $X$ to a $T_0$ space is continuous if and only if it is constant. Between these two cases you have intermediate cases.
Example. Let $X=\{a,b,c\}$ with topology given by $\tau=\{X,\emptyset, \{a\}, \{b,c\}\}$ then a function from $X$ to a $T_0$ space is continuous if and only if $f(b)=f(c)$.