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_{b}:[tex]\Pi[/tex]X

_{a}--> X

_{b}is both continuous and open. I have already done the problem but I would like to check it.

1) Continuity:

Consider an open set U

_{b}in X

_{b}, then P

_{b}

^{-1}(U

_{b}) is an element of the base for the Tychonoff topology on [tex]\Pi[/tex]X

_{a}. Thus, P

_{b}is continuous.

2) Openness:

Let U be an open set in [tex]\Pi[/tex]X

_{a}and let p be a point in U. Then there exists a neighbourhood V of p contained in U. Thus, P

_{b}(p) [tex]\in[/tex] P

_{b}(V) and P

_{b}(V) [tex]\subset[/tex] P

_{b}(U) and every point P

_{b}(p) of P

_{b}(U) has a neighbourhood P

_{b}(V). Thus, P

_{b}(U) is open and P

_{b}is an open map.

I'm less sure about whether my proof of openness is correct. Could you tell me if my answer is correct or not and where I need to improve in my proofs? This question came from General Topology by Stephen Willard (Section 8). Thank you in advance.