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Let $X_1, \dots, X_n$ be topological spaces.

For any $i \in \{1,\dots, n\}$ and any points $x_j \in X_j$, $j\ne i$, show $f: X_i \to X_1 \times \cdots \times X_n$ defined by $f(x)=(x_1, \dots, x_{i-1}, x, x_{i+1}, \dots, x_n)$ is a topological embedding of $X_i$ into the product space.

I need to show that $f$ is injective and continuous and homeomorphic to its image. But I am having trouble showing it is continuous.

Let $U_1\times \cdots \times U_n$ be a basis element of $X_1\times \cdots X_n$ and suppose $x \in f^{-1}(U_1\times \cdots \times U_n)$. Then $f(x)=(x_1, \dots, x_{i-1}, x, x_{i+1}, \dots, x_n) \in U_1\times \cdots \times U_n$.

How can I show this is continuous?

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    $\begingroup$ Have you considered that $f^{-1}(U_1\times\cdots\times U_n)$ equals either $U_i$ or $\emptyset$ and is therefore open by definition? $\endgroup$ – Michael L. May 23 '18 at 7:17
  • $\begingroup$ How do we know it is open by definition? That is what we need to show. $\endgroup$ – user46372819 May 23 '18 at 7:18
  • $\begingroup$ By your choice of basis element, I would say. The definition of an open set in the product topology on $\prod_{i=1}^n X_i$ is a set that can be written as the Cartesian product of open sets in the $X_i$, so if you choose an open set $U_1\times\cdots \times U_{i-1}\times U_i\times U_{i+1}\times \cdots\times U_n$ in the product topology, then $U_i$ should be open by definition. $\endgroup$ – Michael L. May 23 '18 at 7:20
  • $\begingroup$ What is throwing me off is the other coordinates besides $x_i$. I am not sure how to see their preimage. $\endgroup$ – user46372819 May 23 '18 at 7:21
  • $\begingroup$ They are constant. For the basis element you wrote, if $x_j\in U_j$ for all $j\neq i$, then the preimage of the given set is $U_i$. Otherwise, it must be the empty set, as there are no points in the set with the right coordinates. $\endgroup$ – Michael L. May 23 '18 at 7:23
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In general if $f:Y\to Z$ is a function and $Y$ and $Z$ are topological spaces then - if $\mathcal V$ denotes a subbase for the topology on $Z$ - it can be shown that: $$f\text{ is continuous }\iff f^{-1}(\mathcal V)\subseteq\tau_Y\tag1$$where $\tau_Y$ denotes the topology on $Y$. Here $\implies$ is evident.

For any collection $\mathcal W$ of subsets of some set we can agree that $\tau(\mathcal W)$ denotes the smallest topology that contains $\mathcal W$ as a subcollection. Using this notation it can be proved that:$$\tau(f^{-1}(\mathcal V))=f^{-1}(\tau(\mathcal V))\tag2$$

A direct consequence of the RHS of $(1)$ is that $\tau(f^{-1}(\mathcal V))\subseteq\tau_Y$, and applying $(2)$ on the RHS of $(1)$ we find that $$f^{-1}(\tau_Z)=f^{-1}(\tau(\mathcal V))=\tau(f^{-1}(\mathcal V))\subseteq\tau_Y$$

So this states that $f$ is continuous.


This can be used to prove that $f:X_i\to X_1\times\cdots\times X_n$ as described in your question is continuous.

For $\mathcal V$ we take the collection of sets that can be written as $p_j^{-1}(U)$ where $j\in\{1,\dots,n\}$, $p_j$ denotes the projection onto $X_j$ and $U$ is an open subset of $X_j$. This is a suitable subbase for the product topology.

If $j\neq i$ then $f^{-1}(p_j^{-1}(U))=(p_j\circ f)^{-1}(U)\in\{\varnothing, X_i\}\subseteq\tau_{X_i}$. It will take value $\varnothing$ if $x_j\notin U$ and will take value $X_i$ otherwise (note that $p_j\circ f$ is a constant function that takes value $x_j$).

If $j=i$ then $f^{-1}(p_j^{-1}(U))=f^{-1}(p_i^{-1}(U)=(p_i\circ f)^{-1}(U)=U\in\tau_{X_i}$ (note that $p_i\circ f=\mathsf{id}_{X_i}$).

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You have learnt in calculus 102 that a vector valued map $x\mapsto{\bf f}(x)=\bigl(f_1(x),f_2(x),\ldots, f_n(x)\bigr)$ is continuous if all its component maps $f_k$ are continuous. In the case at hand all $f_k$ with $k\ne i$ are constant, and $f_i:\>X_i\to X_i$ is the identity map.

But we can also go back to the neighborhoods. Let a $p\in X_i$ be given and put $$f(p)=(x_1,\ldots, p,\ldots x_n)=:{\bf q}\in X\ .$$ Let an arbitrary neighborhood $V$ of ${\bf q}$ be given. By definition of the product topology on $X$ there is a box neighborhood $B:=V_1\times\ldots\times V_n\subset V$ with $V_k\subset X_k$ a neighborhood of the given $x_k$ when $k\ne i$, and $V_i\subset X_i$ a neighborhood of $p$. It is then obvious that $f^{-1}(B)= V_i$, hence $f(V_i)\subset B\subset V$, as required for continuity of $f$ at the point $p$.

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  • $\begingroup$ It is not entirely obvious that $f^{-1}(B)=V_i$. Intuitively, it is obvious. But the reason it works is because we can write \begin{align*} f^{-1}(B)&=f^{-1}(V_1\times\cdots \times V_{i-1}\times V_i\times V_{i+1}\times \cdots\times V_n)\\&= f_1^{-1}(V_1)\cap \cdots \cap f_{i-1}^{-1}(V_{i-1}) \cap f_i^{-1}(V_i)\cap f_{i+1}^{-1}(V_{i+1}) \cap \dots \cap f_n^{-1}(V_n)\\&=X_i \cap \dots \cap V_i\cap \dots\cap X_i\\&=V_i\end{align*} which is open in $X_i$. $\endgroup$ – user46372819 May 23 '18 at 18:38
  • $\begingroup$ No, the proof really is "obvious" by some pretty fundamental notions from set theory. You can formalize it in the way you've written, but I think most mathematicians would consider this unnecessary in the given setting. There is no need to harp on breaking the function into all its component parts, which just complicates the matter when we consider, for example, an uncountable product of topological spaces. $\endgroup$ – Michael L. May 24 '18 at 7:17

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