Is This Statement in Munkres' Topology Book False?

[facenian]

This exersice is from TOPOLOGY by Munkres and I suspect it is false:
"Let $f:A\rightarrow\prod X_\alpha$ be defined by the equation
$f(a)=(f_\alpha(a))_{\alpha\in J}$;
Let $Z$ denote the subspace $f(A)$ of the product space $\prod X_\alpha$. Show that the image under $f$ of each element of $\mathcal{F}$ is an open set of $Z$."
The topology $\mathcal{F}$ of $A$ is given by the subbasis
$$\delta=\bigcup_{\alpha\in J}\{f_\alpha^{-1}(U):U\in\mathcal{F}_\alpha\}$$
Does anybody has an opinion?

 

[fresh_42]

I think it's true. We have for $U \in \mathcal{F}_{\alpha_0} \, : \,f(f_\alpha^{-1}(U)) \subseteq U \times \prod_{\alpha \neq \alpha_0}f(f^{-1}(X_\alpha))$ and equality on the induced topology of $Z=f(A)$, so $f$ should be open.

But I might have overlooked something - I tend to in topology.

 

[facenian]

Yes the asertion on Munkres book is correct, I was wrong. Thanks