Inverse of a projection function

Question

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?

Answer 1

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.

Answer 2

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.