Let $f: X \to Y$ be a continuous map between topological spaces. I'm trying to show that if $X$ is irreducible, then so is $f(X)$.
Let $Y_1, Y_2 \subset f(X)$ two closed sets for the induced topology from $Y$ on $f(X)$ with the property that
$$f(X) = Y_1 \cup Y_2.$$
By definition of the induced topology, there is $Y_1', Y_2' \subset Y$ closed such that
$$Y_i = Y'_i \cap f(X), \quad i = 1, 2.$$
Now by continuity of $f$, we have that
$$f^{-1}(Y'_i) = f^{-1}(Y_i), \quad i = 1, 2, $$
are closed and that
$$X = f^{-1}(Y'_1) \cup f^{-1}(Y'_2).$$
By using the irreducibility of $X$, we deduce that (without lost of generality)
$$f^{-1}(Y'_1) = \varnothing \quad \text{and} \quad f^{-1}(Y'_2) = X.$$
It follows that
$$Y_1 = Y_1' \cap f(X) = \varnothing \quad \text{and} \quad Y_2 = Y_2' \cap f(X) = f(X),$$
which exactly means that $f(X)$ is irreducible.
Does it seems correct ? I feel like I'm missing something but I do not see what..
Best Answer
Your proof is correct. Maybe spend some time on the final step: if $f^{-1}[Y_2'] = X$ we conclude that $x \in X \to f(x) \in Y_2' \to f(x) \in Y_2$ so $f[X] \subseteq Y_2 \subseteq f[X]$ so indeed $f[X] = Y_2$ and $f^{-1}[Y'_1]=\emptyset$ implies $\emptyset = Y_1$ as otherwise $y \in Y_1 \subseteq f[X] \to (\exists x \in X: f(x)=y) \to (\exists x \in f^{-1}[Y_1'])$ etc. It's a minor point, but might make the final step a bit clearer. Otherwise, no comments.