A topological space X is connected if and only if the group $\operatorname{Hom}(X,\{0,1\})$ is isomorphic to $\mathbb{Z}_2$

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I would like to have some hints for the first part of the following exercise from Kosiowski, which asks to prove a quite general fact

First steps in algebraic topology. Let $X$ be a topological space and define $H(X)$ to be the set of continuous maps from $X$ to $\mathbb Z/2\mathbb Z$ with the discrete topology. If $f,g\in H(X)$, then define $f+g$ for all $x\in X$ by$$(f+g)(x)=f(x)+g(x)\pmod 2$$ Prove that $f+g$ is continuous and $H(X)$ is an abelian group with respect to this operation. Prove that $X$ is connected if and only if $H(X)$ is isomorphic to $\mathbb Z/2\mathbb Z$. Construct examples of topological spaces $X_k$ with $H(X_k)$ isomorphic to $(\mathbb Z/2\mathbb Z)^k$.

Moreover: I know that the set $\{0,1\}$ is a sort of dualizing object, for example inn the context of boolean algebras, and would like to know whether the same intuition about duality could be somehow applied here.

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Continuous maps from $X$ to the two-point discrete space correspond one-to-one with the clopen subsets of $X$. The pointwise sum mod 2 then corresponds to the symmetric difference of two clopen subsets of $X$, which is easily seen to itself be clopen.

By definition, $X$ is connected if and only if it has exactly two clopen sets ($\emptyset$ and $X$), which means equivalently that the group $H(X)$ is cyclic of order 2.

Note that the empty space is not connected (similar to the trivial group not being simple, $1$ not being prime, the trivial ring not being an integral domain or field, the empty $G$-set for a group $G$ not being transitive, etc.). The general situation is referred to as "too simple to be simple" in the nLab.