= Solution
If $f:\widetilde X\to Y$ is continuous, then $f\circ\pi$ is continuous because the quotient map $\pi$ is continuous by definition.
Conversely, suppose $f\circ\pi$ is continuous. For every open set $V\subseteq Y$,
$$
\pi^{-1}(f^{-1}(V))
=(f\circ\pi)^{-1}(V)
$$
is open in $X$. The definition of the quotient topology then says that $f^{-1}(V)$ is open in $\widetilde X$. Hence
$$
\boxed{f\text{ is continuous}\iff f\circ\pi\text{ is continuous}}.
$$
This is the <universal property of the quotient topology>.
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