[proofplan] We prove separately the two requirements for being an [open cover](/page/Open%20Cover) of $X$. First, every member of the pullback family is open in $X$, because it is the preimage of an open subset of $Y$ under the continuous map $f$. Second, every point $x\in X$ is covered: since $\mathcal U$ covers $Y$, the point $f(x)\in Y$ lies in some $U\in\mathcal U$, and therefore $x\in f^{-1}(U)$. [/proofplan] [step:Show that every pulled back set is open in $X$] Define the pullback family \begin{align*} \mathcal V := f^{-1}\mathcal U = \{f^{-1}(U): U\in\mathcal U\}. \end{align*} Let $V\in\mathcal V$. By definition of $\mathcal V$, there exists $U\in\mathcal U$ such that $V=f^{-1}(U)$. Since $\mathcal U\subset\tau_Y$, the set $U$ is open in $Y$. Since $f:X\to Y$ is continuous, the preimage $f^{-1}(U)$ is open in $X$, that is, $f^{-1}(U)\in\tau_X$. Hence $V\in\tau_X$. Therefore every member of $\mathcal V$ is open in $X$. [/step] [step:Show that the pulled back family covers every point of $X$] Let $x\in X$. Since $f:X\to Y$, we have $f(x)\in Y$. Because $\mathcal U$ is an open cover of $Y$, there exists $U_x\in\mathcal U$ such that $f(x)\in U_x$. By definition of preimage, \begin{align*} x\in f^{-1}(U_x). \end{align*} Since $U_x\in\mathcal U$, the set $f^{-1}(U_x)$ belongs to $\mathcal V$. Thus $x$ lies in some member of $\mathcal V$. [guided] We now verify the covering property point by point. The goal is to prove that every $x\in X$ belongs to at least one set in the family \begin{align*} \mathcal V := \{f^{-1}(U):U\in\mathcal U\}. \end{align*} Take an arbitrary point $x\in X$. Because $f$ is a map from $X$ to $Y$, the value $f(x)$ is a point of $Y$. The hypothesis that $\mathcal U$ is an open cover of $Y$ means two things: each member of $\mathcal U$ is open in $Y$, and every point of $Y$ belongs to at least one member of $\mathcal U$. Applying the second part to the point $f(x)\in Y$, there exists a set $U_x\in\mathcal U$ such that \begin{align*} f(x)\in U_x. \end{align*} Now we translate this membership statement back to $X$ using the definition of preimage. For a subset $U_x\subset Y$, the preimage $f^{-1}(U_x)$ is the subset of $X$ consisting exactly of those points whose image under $f$ lies in $U_x$. Since $f(x)\in U_x$, we get \begin{align*} x\in f^{-1}(U_x). \end{align*} Finally, because $U_x\in\mathcal U$, the set $f^{-1}(U_x)$ is one of the members of $\mathcal V$. Therefore the arbitrary point $x\in X$ lies in a member of $\mathcal V$. [/guided] [/step] [step:Conclude that the pullback family is an open cover of $X$] The first step proved that $\mathcal V\subset\tau_X$. The second step proved that for every $x\in X$, there exists $V\in\mathcal V$ such that $x\in V$. Equivalently, \begin{align*} X\subset \bigcup_{V\in\mathcal V} V. \end{align*} Since every $V\in\mathcal V$ is a subset of $X$, also \begin{align*} \bigcup_{V\in\mathcal V} V\subset X. \end{align*} Therefore \begin{align*} X=\bigcup_{V\in\mathcal V} V. \end{align*} Thus $\mathcal V=f^{-1}\mathcal U$ is an open cover of $X$. [/step]