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A map $f: X \to Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show that $ \pi_1: X \times Y \to X$ is an open map.

I say:

Let there exist $\lambda \in X \times Y$ such that $\lambda$ is open and $f(\lambda) \in Y$. If $\lambda$ is open in $ X \times Y$ then it is an element of $ U \times V \in X \times Y$ such that $ U \in X$ and $V \in Y$ are open sets. Now, if $\lambda$ is $\in U \times V$ then this map must be sending some $\lambda$ to $V$ in $X$. I can't say why I made the last assumption, it just seemed intuitively right? How do I show it, though?

But I suspect that showing that the whole $X$ is an open set (i.e. if we were discussing topologies) would be much easier way to show that this is an open map but I can't for the life of me figure out how to do that.

Siyanda
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It is enough only to consider a basis of the product topology $X \times Y$, which is $\{(U,V)\mid U\text{ is open in }X, V\text{ is open in Y}\}$. Since $\pi_1(U,V)=U$ is open, it is quite obvious that $\pi_1$ is an open map...

Henry
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