Show that any continuous map $f: X \rightarrow Y$ induces a map of semisimplicial sets $Sing(X). \rightarrow Sing(Y).$

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I want to show that any continuous map $f: X \rightarrow Y$ induces a map of semisimplicial sets $Sing(X). \rightarrow Sing(Y).$, but I'm confused about how to do so. I guess the main confusion is that I'm not really sure of how $Sing(X).$ and $Sing(Y)$ might "talk" to each other at all. Presumably we want to compose with $f$.

Moreover, doesn't this map of semisimplicial sets have the structure of a natural transformation?

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Assuming by semisimplicial set you mean a simplicial set without degeneracies (though, in truth, this works for simplicial sets as well, and other things):

$f$ induces a map $\mathrm{Sing}(X) \to \mathrm{Sing}(Y)$ by composition, just as you stated. You'd need to check that these diagrams commute for face maps $\delta_i : \Delta^n \to \Delta^{n+1}$:

$$ \begin{array}{ccc} \mathrm{Map}(\Delta^{n+1},X) & \xrightarrow{\delta_i^*} & \mathrm{Map}(\Delta^n, X) \\ f_* \downarrow & & f_* \downarrow \\ \mathrm{Map}(\Delta^{n+1},Y) & \xrightarrow{\delta_i^*} & \mathrm{Map}(\Delta^n, Y) \end{array} $$

But that's immediate; $\delta_i^*$ is given by precomposition, and $f_*$ by postcomposition. More specifically, if $\varphi \in \mathrm{Map}(\Delta^{n+1},X)$, then $f_*(\delta_i)^* \varphi = f \circ \varphi \circ \delta_i = \delta_i^*f_* \varphi$.