Let $(X,\tau)$ and $(Y,\tau_1)$ be topological spaces and $f:(X,\tau) \rightarrow (Y,\tau_1)$ be a continuous open mapping. If $(X,\tau)$ is a Baire space, prove that $(Y,\tau_1)$ is a Baire space.
My Proof is that Let $\{X_n\}^\infty_{n=1}$ be a sequence of open dense sets in $X$. Since $X$ is a Baire space, $\bigcap_{n=1}^\infty X_n$ is dense. This implies $\left(\bigcap_{n=1}^\infty X_n\right) \cap U \ne \emptyset$ where $U$ is any open set in $X$. Let $x\in \left(\bigcap_{n=1}^\infty X_n\right) \cap U$, then It would $x$ should be in all $X_n$ and $U$; thus $f(x) \in \left(\bigcap_n^\infty f(X_n)\right)\cap f(U)$. $f(X_n)$ and $f(U)$ is open in $Y$ since $f$ is continuous open mapping function. This holds for all $x$ and $U$. Hence $Y$ is a Baire space.
In this proof, I assume that $f(X_n)$ is an open dense set in $Y$. Can I do that? if so, why am I allowed to do this?
If anything wrong, please correct me. Thanks
Let $\{Y_n\}^\infty_{n=1}$ be a sequence of open dense sets in $Y$. If $(Y,\tau_1)$ is not a Baire space, then there is an open set $V\subset Y$ that $\left(\bigcap_{n=1}^\infty Y_n\right) \cap V =\varnothing$. So there is $$ f^{-1}\left(\bigcap_{n=1}^\infty Y_n\right) \cap f^{-1}(V) =\bigcap_{n=1}^\infty f^{-1}(Y_n) \cap f^{-1}(V) =f^{-1}(\varnothing)=\varnothing\tag1 $$ Since $f$ is continuous open, $f^{-1}$ is continuous. By this post, there is $$ f^{-1}(\overline S) \subset \overline {f^{-1}(S)} $$ Let $f^{-1}(Y_n)=X_n$. Then $\overline{Y_n}=Y$ for $Y_n$ is dense in $Y$. Assume $f$ be surjective. Then $X=f^{-1}(Y)$. Hence $$ X=f^{-1}(Y)=f^{-1}(\overline{Y_n})\subset \overline {f^{-1}(Y_n)}\subset \overline{X_n} $$ Thus $\overline{X_n}=X$, i.e. $X_n$ is dense in $X$.
Let $U=f^{-1}(V)$. Since $f$ is continuous, $U$ is open. Also $X_n$ is open for $Y_n$ is open. So $(1)$ becomes $$ \bigcap_{n=1}^\infty X_n \cap U=\varnothing $$ which contradict $X$ is Baire space.