# Inverse of a projection function

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?

• 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? Commented Jun 25, 2019 at 18:49
• @Arturo Magidin Commented Jun 25, 2019 at 21:41

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.
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.
• 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. Commented Jun 26, 2019 at 21:03