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.


1 Answer 1


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.


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