# Proof that direct product of connected spaces is connected

I am trying to understand a piece of the proof that the direct product of connected spaces is itself a connected space, as given by Lee in "Introduction to Topological Manifolds". By induction, we may consider the case with two spaces.

So suppose that $$X$$ and $$Y$$ are connected topological spaces, and suppose for sake of contradiction that there exist open sets $$U$$ and $$V$$ that disconnect $$X\times Y$$. Let $$(x_0,y_0)$$ be a point in $$Y$$.

The next part is where I am confused. The author writes "The set $$\{x_0\}\times Y$$ is connected because it is homeomorphic to $$Y$$. Why is This space homeomorphic to $$Y$$? I don't see how there can be an injective map from $$\{x_0\}\times Y$$ to $$Y$$. What am I missing? Is there some obvious homeomorphism?

• The obvious map from $\{x_0\} \times \mathrm{Y}$ to $\mathrm{Y}$ is $(x_0, y) \mapsto y,$ which, is inverse of $y \mapsto (x_0, y).$ – Will M. Jan 10 '19 at 20:00
• $f(x_0,y)=y$ is injective – Randall Jan 10 '19 at 20:00
• You're likely missing that $x_0$ is fixed. – Randall Jan 10 '19 at 20:00

Take the map $$\phi : \{x_0\} \times Y \to Y$$ that maps $$(x_0, y) \mapsto y$$. It's clear that this is a bijective continuous function. Furthermore, it's inverse $$\phi^{-1}$$ is the map which takes $$y \mapsto(x_0, y)$$ is also clearly continuous, which tells us that $$\phi$$ is an open mapping. Thus we have a homeomorphism.