Proposition 1.2.1.11. Let $f: S_{} \rightarrow T_{}$ be a map of simplicial sets, and suppose that $S_{}$ is connected. Then there is a unique connected component $T'_{} \subseteq T_{}$ such that $f( S_{} ) \subseteq T'_{}$.
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Proof. Let $T'_{}$ be the smallest summand of $T_{}$ which contains the image of $f$ (the existence of $T'_{}$ follows from Remark 1.2.1.3: we can take $T'_{}$ to be the intersection of all those summands of $T_{}$ which contain the image of $f$). We will complete the proof by showing that $T'_{}$ is connected. Since $S_{}$ is nonempty, $T'_{}$ must be nonempty. Let $T''_{} \subseteq T'_{}$ be a summand; we wish to show that $T''_{} = T'_{}$ or $T''_{} = \emptyset $. Note that $f^{-1}(T''_{} )$ is a summand of $S_{}$ (Remark 1.2.1.5). Since $S_{}$ is connected, we must have $f^{-1}( T''_{} ) = S_{}$ or $f^{-1}( T''_{} ) = \emptyset $. Replacing $T''_{}$ by its complement if necessary, we may assume that $f^{-1}( T''_{} ) = S_{}$, so that $f$ factors through $T''_{}$. Since $T''_{}$ is a summand of $T_{}$ (Remark 1.2.1.4), the minimality of $T'_{}$ guarantees that $T''_{} = T'_{}$, as desired. $\square$