Partial mapping, how does the inverse look like?

Question

Let $X_1,X_2,Y$ be topological spaces. Let $f:X_1\times X_2 \to Y$ be continous at $a=(a_1,a_2)$. Show that the partial mappings $f_1:X_1\to Y; x\mapsto f_1(x) = f(x,a_2)$ is continous at $a_1$ and $f_2: X_2\to Y; x\mapsto f_2(x) = f(a_1,x)$ is continous at $a_2$.

Looks like a straight forward exercise, but I get a bit confused by some details. Here's what I've got:

Let $U$ be a neighbourhood of $f_1(a_1) = f(a_1,a_2)$, I want to show $f^{-1}_1(U)$ is a neighbourhood of $a_1$, or in other words. I want to show existance of a $W\in \tau_{X1}$ such that $a_1\in W\subseteq f^{-1}_1(U)$

Since $f$ is continous at $a$, I know $f^{-1}(U)$ is a neighbourhood of $(a_1,a_2)$. Which translates into $(\exists V\in \tau_{X_1\times X_2})(a\in V\subseteq f^{-1}(U))$.

By definition of the producttopology $\tau_{X_1\times X_2}$ there exist $V_1\times V_2$ where $V_1\in \tau{X_1}$ and $V_2\in \tau{X_2}$ such that $(a_1,a_2)\in V_1\times V_2\subseteq V\subseteq f^{-1}(U)$.

I guess this $W=V_1$ would be a good choice but I don't see why $V_1\subseteq f^{-1}_1(U)$. I don't think $f^{-1}_1(U)\times f^{-1}_2(U) \subseteq f^{-1}(U)$. (see image below, is this visual correct?)

I'm searching for a connection between $f^{-1}(U)$ and $f^{-1}_1(U)$.

Answer 1

Let $v\in V_{1}$.

Then $\left(v,a_{2}\right)\in V_{1}\times V_{2}\subseteq f^{-1}\left(U\right)$ so that $f_{1}\left(v\right)=f\left(v,a_{2}\right)\in U$.

The last statement is the same as $v\in f_{1}^{-1}\left(U\right)$ and proved is now that $V_{1}\subseteq f_{1}^{-1}\left(U\right)$.