I am answering the question:
Let $X = \{1, 2, 3, 4, 5\}$ with topology $\{\emptyset, X, \{1\}, \{3, 4\}, \{1, 3, 4\}\}$, and let $Y = \{A, B\}$ with topology $\{\emptyset, Y, \{A\}\}$. Find all continuous functions from $X$ to $Y$. How many functions from $X$ to $Y$ are there altogether?
This is what I have thus far:
Let $(X,\tau)$ and $(Y,\tau^\prime)$ be topological spaces. A function from $X$ to $Y$ is said to be continuous if given any open subset $U$ of $Y$, then $f^{-1}(U)$ is an open subset of $X$.
a.
Continuous functions from $X$ to $Y$ can be characterized as those that map open subsets of $X$ to open subsets to $Y$. In total, there exist $8$ continuous functions from $X$ to $Y$:
\begin{align*}
f&: \{{1}\} \rightarrow A\\
f&: \{{3, 4}\} \rightarrow A \\
f&: \{{1, 3, 4}\} \rightarrow A \\
f&: X \rightarrow A \\
f&: \{{1}\} \rightarrow Y \\
f&: \{{3, 4}\} \rightarrow Y \\
f&: \{{1, 3, 4}\} \rightarrow Y \\
f&: X \rightarrow Y
\end{align*}
My question is: how do I handle the empty set? I know that if I am mapping to or from the empty set, $X \times Y = \emptyset$, i.e. $\emptyset$ is the only subset of $X \times Y$. So, we shouldn't be able to map $\emptyset$ to a nonempty set, correct? And, should we be able to map a nonempty set to $\emptyset$? Thanks for the help y'all!
Best Answer
The fact that the preimage of an open is open IS NOT equivalent to the fact that $f$ maps open sets to open sets. Such maps are called open maps. There exist continuous maps that are not open, open maps that are not continuous, open continuous maps and maps which are neither continuous nor open!
To find continuous maps from $X$ to $Y$ in your case amounts exactly to find all the possible preimages of $A$. Since $Y$ has only two elements, the preimage of $A$ determines uniquely the map. So there are exactly 5 continuous maps $X\to Y$. Two of them are the constant ones, the third is the one mapping $1$ to $A$ (and everything else to $B$), the fourth is the one mapping $3$ and $4$ to $A$ and the fifth maps $1$,$3$ and $4$ to $A$.
You don't have to care about $\varnothing$, because its preimage under any map is again $\varnothing$ which has to be open by definition of topology. In the same way, you don't have to care about the preimage of $Y$, which is always $X$.