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

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 $f_*(\delta_i)^* \varphi = f \circ \varphi \circ \delta_i = \delta_i^*f_* \varphi$.
