Given the following defintion of a $T_{3 \frac{1}{2}}$ Space:
$\underline{Z}$ is called a $T_{3 \frac{1}{2}}$ Space if for every point $z \in Z$ and every set $A \subset Z$, with $z \not \in A$ there exists a continuous map $m: \underline X \rightarrow \underline{[0,1]}$ with $m(z) = 0$ and $m[A] \subset \{1\}$.
In the chapter about $T_{3 \frac{1}{2}}$ Spaces the author defines $\underline Y$ as a s.c. $T_{3 \frac{1}{2}}$-Reflection of a space $\underline X = (X, T_x) $ like this:
$S = \{f^{-1} [B] \ |\ f : \underline X → \underline {[0,1]} \text{ continuous map }, B \in T_{[0,1]} \}$
$\underline Y = (X, T_y), \text{where} \ T_y$ is a Topology generated by the Subbasis $S$.
If my understanding is right, then he just defines an Initial Topology on a set of all continuous maps from $\underline X$ to $\underline{[0,1]}$ and calls it $\underline Y$.
What I don't understand is the following. He proceeds by stating that it is "obvious" that for some $f$ it holds that:
$f: \underline Y \rightarrow \underline{[0,1]}$ is continuous $ \iff $ $f: \underline X \rightarrow \underline{[0,1]}$ is continuous.
While I easily get why the $\Leftarrow$ holds, I just can't understand why $\Rightarrow$ is supposed to hold.
If $f: \underline Y \rightarrow \underline{[0,1]}$ is continuous, then why can't it be that $f: \underline X \rightarrow \underline{[0,1]}$ is a discontinuous map?
The identity $1_{X,Y}: \underline{X} \to \underline{Y}$ is continuous because the subbase $\mathcal{S}$ is by definition is subfamily of $\mathcal{T}_X$, so that the topology on $\underline{Y}$ is coarser than $\mathcal{T}_X$.
So if $f: \underline{Y} \to [0,1]$ is continuous, the map $f: \underline{X} \to [0,1]$ (same function) can be seen as $f \circ 1_{X,Y}$ and is continuous as a composition of continuous maps.