# Proofs inolving mapping from countably infinite product topology to another countably infinite product topology

Proposition: Let($$X_i,\tau_i$$) and ($$Y_i,\tau^{‘}_i$$), i∈N,be countably infinite families of topological spaces having product spaces( $$\Pi X_i,\tau$$)and( $$\Pi Y_i, \tau_{‘}$$), respectively.If the mapping $$h_i$$: ($$X_i,\tau_i$$)$$\mapsto$$($$Y_i,\tau^{′}_i$$ ) is continuous for each i∈N, then so is the mapping h: (∏Xi,τ)$$\mapsto$$(∏Yi,τ′)given by h:($$\Pi x_i$$)=$$\Pi h_i(x_i)$$ that is,$$h(x_1,x_2,...,x_n,…,)$$=$$〈h_1(x_1),h_2(x_2),...,h_n(x_n),...〉$$

Note: we are working with the countable infinite product topology

If in the above proposition each mapping $$h_i$$ is also(a) one-to-one,(b) onto,(c) onto and open,(d) a homeomorphism, prove that h is respectively (a) one-to-one, (b) onto, (c) onto and open, (d) a homeomorphism

a)injective just means if f(a)=f(b) then a=b, so assume h(x)=h(y) where x and y are infinite tuples. So $$(h_1(x_1),h_1(x_2),h_1(x_3)....)$$=$$(h_1(y_1),h_2(y_2),h_3(y_3)....)$$. Now since each $$h_i$$ are one to one, we have $$h_1(x_i)=h_1(y_1)$$ so $$x_1=y_1$$ and this is true for all $$x_i$$and $$y_i$$ so x=y. Thus h is injective.

b)onto means if y is in the codomain, there exist an x such that f(x)=y. If each $$h_i$$ is onto, then for all infinite n tuple ($$y_1,y_2,y_3,y_4....$$) there exist a infinite x tuple ($$x_1,x_2,x_3,x_4)$$ such that ($$h_1(x_1),h_1(x_2),h_1(x_3)....)$$ =($$y_1,y_2,y_3,y_4... )$$ By our definition of h, h is onto.

c) Now we assume each $$h_i$$ is onto and map open sets onto open sets. We have already shown that h is onto if each $$h_i$$ is onto. Take a basic open set in the product topology which is of the form $$O_1\times O_2 \times O_3 \times R \times R \times R$$ with all but finitely many tuples are Rs. Now applying h we get, ($$h_1(O_1),h_2(O_2),h_3(O_3),h_4(R)....).$$ Since each $$h_i$$ are open mapping, each component in their respective codomain is open so h is open.

d)The definition of homeomorphism I will use is bijective, open, and continuous mapping and from the proposition above, continuity is already proven. Thus combining the four properties. if each hi is homeomorphic, then h is also homeomorphic.

In fact, the cardinality of the index size of the product is irrelevant. It could be countable, uncountable or finite. The proofs are exactly the same.

You could just write $$h: X:=\prod_{i \in I} X_i \to Y:=\prod_{i \in I} Y_i$$ and prove the first continuity fact by noting $$\pi'_i \circ h = h_i \circ \pi_i$$ for all $$i$$ and so by the universal property of $$(Y, \pi'_i, i \in I)$$, $$h$$ is continuous. This makes for smoother notation (moreover one should use MathJax on this site for clarity's and readability's sake)

The injectivity and surjectivity facts are obvious as you noted.

$$h(x)=h(y) \iff \forall i: \pi'_i(h(x)=\pi'_i(h(y) \iff \forall i: h_i(\pi_i(x)) = h_i(\pi_i(y))$$ and as all $$h_i$$ are 1-1 we continue with $$\iff \forall i: \pi_i(x)=\pi_i(y) \iff x=y$$.

If $$y\in Y$$, for each $$i$$ we pick $$x_i \in X_i$$ so that $$h_i(x_i) = \pi'_i(y)=y_i$$ and then clearly $$h(x)=y$$ for $$x= (x_i)_{i \in I}$$.

It suffieces to check openness on basic open sets (a general proposition) and if $$O = \prod_i O_i$$ is basic open in $$X$$ (depending on some finite $$F \subseteq I$$, say), $$h[O] = \prod_{i \in I} h_i[O_i]$$ is of the same form, as $$h_i[O_i]= h_i[X_i]=Y_i$$ for all $$i \notin F$$; this is where ontoness of the $$h_i$$ is used, so $$h[O]$$ is also (basic) open.

As a homeomorphism is 1-1 open and onto, the last follows immediately from the previous three.

This is all the proof comes down to, really.