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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?

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

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