How do 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 continuous?

Question

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?

Answer 1

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.

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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 $p_i\circ f=\mathsf{id}_{X_i}$).

Answer 2

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 $p$.