homotopy between two functions

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Let us discuss this problem:

Let $A=\{a_{1},a_{2},\ldots,a_{n}\}$, $B=\{b_{1},b_{2},\ldots,b_{n}\}$ and $C=\{c_{1},c_{2},\ldots,c_{n-1}\}$ be discrete finite sets embedded in a unit sphere $S^2=x^2+y^2+z^2={1}$ so that the topologies of $A$, $B$ and $C$ are induced topology from the topology of the sphere (Note that the topologies of $A$, $B$ and $C$ are $P(A)$, $P(B)$ and $P(C)$ respectively, where $P(A)$ denotes the power set of $A$). Assume also that the points of $B$ are imbedded in a great circle in $S^{2}$ such that the arc length between each two of these points is equal. Also suppose that elements of $C$ are imbedded in a great circle (it can be the same great circle of $B$) so that the arc length between each two of these elements is equal.

Define $f_{0}\colon A \rightarrow B$ by $f_{0}(a_{i})=b_{i}$ for $i=1,2,\ldots,n$. Define also a function $f_{1}\colon A \rightarrow C$ by $f_{1}(a_{1})=f_{1}(a_{n})=c_{1}$ and $f_{1}(a_i)=c_{i}$ otherwise.

Question Is it possible to define a homotopy $F\colon A \times [0,1] \rightarrow S^{2}$ between $f_{0}$ and $f_{1}$?

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$X$ and $Y$ are not continuous functions. How does the notion of homotopy apply here?

If, on the other hand, you want to ask if $X$ and $Y$ are homotopy equivalent, then this is clearly not possible either as the cardinality of path components of a space is a homotopy invariant and $X$ and $Y$ have distinct numbers of path components.

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I'll add another answer as this is now a new (and well-defined) question.

Note that any map to $S^2$ which is not surjective is homotopic to the constant map$^{[1]}$ (because $S^2\setminus\{p\}$ is contractible) and so, as homotopy is an equivalence relation on the set of maps between $A$ and $S^2$, and $f_0$ and $f_1$ are both non-surjective (by cardinality arguments), it follows that $f_0$ and $f_1$ are homotopic.


$[1]$I say the constant map here as I'll assume that we're working with pointed maps for convenience. The above argument still works for non-pointed maps but you have to be a bit more careful about which constant map you are homotoping to and need to also show there is a homotopy between these two constant maps (which is easy enough because $S^2$ is path connected.)