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I have got the following question;

Let $(X_1, \tau_1), (X_2, \tau_2)$ be topological spaces and $X = X_1 \times X_2$. Equip $X$ with the product topology $\tau$ so that, by definition of $\tau$, the projection function $\pi_i : X \rightarrow X_i$ defined by \begin{align} \pi_i(x) = x_i \quad \text{for} \ x = (x_1, x_2) \in X, \ \forall i \in I \end{align} is a continuous function. In general, the inverse image of this projection function, $\pi_i^{-1} : X_i \rightarrow X$, is not a well defined function. However, if we consider the special case where $X_2$ is a singleton set, i.e. $X_2 = \{x_2\}$, then $\pi_1^{-1}$ is actually a well-defined function.

Is it true that this function $\pi_1^{-1}$ is a continuous function? If true, how can I show that?

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  • $\begingroup$ Can you prove that in general, for a fixed $z\in X_2$, the map $f\colon X_1\to X_1\times X_2$ given by $f(x) = (x,z)$ is a homeomorphism onto its image? $\endgroup$ Commented Jun 25, 2019 at 18:49
  • $\begingroup$ @Arturo Magidin $\endgroup$
    – F.Vitiello
    Commented Jun 25, 2019 at 21:41

2 Answers 2

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If we have a set $A\subseteq X_1\times X_2$ such that $\pi_1 \restriction_A$ is injective (and onto), then this is a homeomorphism, due to the projection being an open map.

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The function $f$ is clearly a bijection.

In order to show that $f^{-1}$ is continuous consider $U_1 \in \tau_1$, \begin{align} f(U_1) &= \{(x_1, z) \in X_1 \times \{z\} \ \mathrm{s.t.} \ f(x_1) = (x_1, z) \ \mathrm{for} \ x_1 \in U_1 \} \\ &= \pi_1^{-1}(U_1) \in \tau \ \text{by the definition of the product topology} \end{align}

Similarly, in order to show that $f$ is continuous consider the set $O \in \tau$. This set $O$ is of the form $O_1 \times \{z\}$, so \begin{align} f^{-1}(O) &= f^{-1}(O_1 \times \{z\}) \\ &= \{x_1 \in X_1 \ \mathrm{s.t.} \ f(x_1) = (x_1, z) \ \mathrm{for} \ (x_1, z) \in O_1 \times \{z\} \} \\ &= \pi_1(O_1 \times \{z\}) \\ &= O_1 \end{align}

So in order to show that $f$ is a homeomorphism I need to conclude that $O_1 \in \tau_1$, but I do not know how to conclude that.

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    $\begingroup$ If $f(x)=(x,z)$ as Arturo suggested, then $f^{-1}[(O_1 \times O_2) \cap (X_1 \times \{z\})] = O_1 \in \tau_1$ if $z \in O_2$, otherwise empty; and sets of that form form a base for the subspace $X_1 \times \{z\}$ so $f$ is continuous. OTOH, if $O$ is open in $X_1$, $f[O_1] = (O_1 \times X_2) \cap (X_1 \times \{z\})$ so also basic open and $f$ is an open map. Done. $\endgroup$ Commented Jun 26, 2019 at 21:03

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