Infinite Join homeomorphism property preserved

Question

Problem:

In the lecture notes the following is writen:

Let $X_1,X_2,X_3,...$ be topological spaces. Put $X=\star_{i=1}^{\infty} X_i$ (join). Then assume $f_i:X_i\approx Y_i$ is a homeomorphism for each $i$, where $Y_i$ is a topological space. To showthat $\star X_i\approx \star Y_i$ (homeomorphic), define $f:\star X_i \rightarrow \star Y_i$ by $f(t_mx_m)=(t_mf(x_m))$ and clearly has inverse $f^{-1}(t_iy_i)=(t_if^{-1}(y_i))$. So $f$ is a homeomorphism.

Is this true? I fail to see why $f$ is continuos. Is the topology just induced from product topology?

Answer 1

What is written does not make sense. $\star X_i$ does not make sense, $f:\star X_i\to\star Y_i$ make less sense (we would expect the map to depend on the index $i$). I assume by $\star X_i$ you mean $X$, i.e. $\star_{i=1}^\infty X_i$, but $\star X_i$ is sloppy notation. I also think $f(t_mx_m):=t_mf(x_m)$ is a bit of a terse definition... and it should at least read $t_m f_m(x_m)$.

In this, I use the following model of finite joins:

If $A_1,A_2,\cdots,A_n$ are given spaces, we define $A_1\star A_2\star\cdots\star A_n$ to be the quotient of $\Delta^{n-1}\times\prod_{j=1}^nA_j$ by the relations which identify $(x,a_1,\cdots,a_j,\cdots,a_n)$ with $(x,a_1,\cdots,a'_j,\cdots,a_n)$ if the $j$th coordinate of $x$ is zero.

You say you define $\star_{i\in I}X_i$ as the direct limit of the finite joins $\star_{i\in F}X_i$, $F\subseteq I$ finite. I assume all the $X_\bullet$ are nonempty. I would assume the arrows in that diagram run $\star_{i\in F}X_i\to\star_{i\in F'}X_i$ for $F\subseteq F'$ and are defined via the over-maps $\Delta^{|F|-1}\times\prod_{j\in F}X_j\to\Delta^{|F'|-1}\times\prod_{j\in F'}X_j$ which includes the simplex into the faces corresponding to $F\subseteq F'$ (to make it super precise we should fix a numbering on $F,F'$, but that would clutter the notation). The maps into $\prod_{j\in F'\setminus F}X_j$ can be chosen to be constant to fixed points of each $X_j$, it does not really matter since these components will vanish in the quotient anyway.

So we put: $$X:=\varinjlim_{F\subseteq I,\,F\text{ finite}}\star_{j\in F}X_j$$

We are given a family of nonempty spaces $(Y_i)_{i\in I}$ and homeomorphisms $f_i:X_i\cong Y_i$ for each $i$, and similary define $Y:=\varinjlim_{F\subseteq I,\,F\text{ finite}}\star_{j\in F}Y_j$. We can define homeomorphisms $\star_{i\in F}X_i\cong\star_{i\in F}Y_i$, $F$ finite, through the over-maps: $$\Delta^{|F|-1}\times\prod_{i\in F}X_i\overset{1\times\prod_{i\in F}f_i}{\cong}\Delta^{|F|-1}\times\prod_{i\in F}Y_i$$And then it's easy to see the homeomorphisms $\star_{i\in F}X_i\cong\star_{i\in F}Y_i$ assemble to a natural isomorphism of diagrams. Hence we find a canonical isomorphism - because $\varinjlim$ is a functor... - $X\cong Y$.

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Because the inner square commutes, by the universal property of quotients we can convince ourselves the rest of the diagram exists and commutes. That gives naturality in the homeomorphisms $\star_{j\in F}X_j\cong\star_{j\in F}Y_j$.