Continuity in topology I came across an exercise problem which says:
Let $X=\{1,2,3\}$ with topology $T= \{\{1,2\}, \{1,2,3\}, \{3\}, \emptyset\}$ and $Y=\{1,2\}$ with topology $t=\{\{1\},\{2\},\{1,2\},\emptyset\}$. When is $F:X\rightarrow Y$ continuous with respect to these topologies? 
My question is how can you possibly generate a map between these sets without the actual definition. For e.g Let $f:X\rightarrow Y$ be defined by f(x)=x and then we can check for continuity. I know the definition of continuous functions is different in topology but still.
 A: HINT: $F:X\to Y$ is continuous if $F^{-1}[U]$ is open in $X$ whenever $U$ is open in $Y$. In other words, in your case $F:X\to Y$ is continuous if $F^{-1}[U]\in T$ for each $U\in t$. Now $F^{-1}[\{1,2\}]=\{1,2,3\}\in T$ no matter what $F$ is, and $F^{-1}[\varnothing]=\varnothing\in T$ no matter what $F$ is, so it’s only the sets $\{1\}$ and $\{2\}$ in $t$ that might cause trouble. You need to choose $F$ so that $F^{-1}[\{1\}]$ and $F^{-1}[\{2\}]$ are members of $T$. Moreover, $F$ sends every element of $X$ either to $1$ or to $2$, so every element of $X$ must belong to exactly one of $F^{-1}[\{1\}]$ and $F^{-1}[\{2\}]$. There are only a few ways to split $X$ into two disjoint open sets. One of them is to split it as $X\cup\varnothing$; you could have $F^{-1}[\{1\}]=X$ and $F^{-1}[\{2\}]=\varnothing$, meaning that $F$ is the constant function on $X$ that sends every member of $X$ to $1$. 


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*What other continuous $F:X\to Y$ corresponds to this splitting of $X$?  

*How else can you split $X$ into two open parts, and what continuous $F:X\to Y$ do you then get?

