Is the subbasis of a topological space preserved under an open continuous map? Let $X$ and $Y$ be topological spaces, and $f:X\to Y$ a continuous, surjective, and open function. If $\{S_\alpha\}_\alpha$ is a subbasis of $X$, is $\{f(S_\alpha)\}_\alpha$ a subbasis of $Y$ ? That is, if the topology on $X$ is the smallest one containing $\{S_\alpha\}_\alpha$, is the topology on $Y$ the smallest one containing $\{f(S_\alpha)\}_\alpha$ ? Note that the image of a basis of $X$ is a basis of the topology on $Y$.
 A: The answer is NO: Let
$$f:\mathbb{Z}\to\{0,1\}$$
be a function defined by $f(\mbox{even})=0$ and $f(\mbox{odd})=1$.
Give $\mathbb{Z}$ and $\{0,1\}$ the discrete topologies. Then $f$ is open and continuous. The rays in $\mathbb{Z}$  of the form
$$(n,\infty)\qquad\mbox{and}\qquad(-\infty,n)$$
form a subbasis of $\mathbb{Z}$. Their images in $\{0,1\}$ are all the entire set $\{0,1\}$, so they do not form a subbasis of the discrete topology on $\{0,1\}$.
A: I think yes. Let's explore!

Let $f \colon X \to Y$ be an open continuous map, where $X$ and $Y$ are topological spaces. Let $\mathscr{S}$ be a subbasis for $X$. Then the collection
  $$ \big\{ \, f(S) \, \colon \, S \in \mathscr{S} \, \big\} \tag{1} $$
  is a subbasis for the subspace $F(X)$ of $Y$. [Does this conclusion hold?]

In what follows, I'll be using the approach used by Munkres in defining subbasis for a topological spaces.
Proof:

First, we note that, as 
  $$ \bigcup_{S \in \mathscr{S}} S = X, $$
  so we must also have
  $$ \bigcup_{S \in \mathscr{S}} f(S) = f \big( \bigcup_{S \in \mathscr{S}} S  \big) = f(X). $$
  Thus the collection (1) is a subbasis for some topology on $f(X)$. 
For any $n \in \mathbb{N}$ and for any sets $S_1, \ldots, S_n \in \mathscr{S}$, the set 
  $S_1 \cap \cdots \cap S_n$ is a basis set for the topology of $f(X)$, and conversely any basis set for the topology of $X$ is of the form $S_1 \cap \cdots \cap S_n$, where $n \in \mathbb{N}$ and $S_1, \ldots, S_n \in \mathscr{S}$.
Let $V$ be any open set in $f(X)$. Then $V = f(X) \cap V^\prime$ for some open set $V^\prime$ in $Y$. Then we find that
  $$
\begin{align} &= f^{-1} \big( f(X) \cap V^\prime \big) \\
&= f^{-1} \big( f(X) \big) \cap f^{-1} \big( V^\prime \big) \\
&= X \cap f^{-1} \big( V^\prime \big) \\
&= f^{-1} \big( V^\prime \big),
\end{align}
$$
  which is an open set in $X$, being the inverse image under the continuous map $f$ of an open set $V^\prime$ in $Y$. 
In short, for any open set $V$ in $f(X)$, the inverse image $f^{-1}(V)$ is again an open set in $X$ and thus expressable as a union of sets of the form $S_1 \cap \cdots \cap S_n$, where $n \in \mathbb{N}$ and $S_1, \ldots, S_n \in \mathscr{S}$.
However, there is one problem: Unless $f$ is injective, the equality
  $$ f \big( S_1 \cap \cdots \cap S_n \big) = f \big( S_1 \big) \cap \cdots \cap f \big( S_n \big) $$
  need not hold.
Moreover, although
  $$ f\big( f^{-1}(V) \big) \subset V $$
  holds, the equality
  $$ f\big( f^{-1}(V) \big) = V $$
  does not hold in general if $f$ is not surjective.

Please refer to Sec. 2 in Munkres's Topology, 2nd edition.
Can you figure out the answer to your question from this?
