Vertical Section of Continuous Function is Continuous
Theorem
Let $X$, $Y$ and $T$ be topological spaces.
Equip the Cartesian product $X \times Y$ with the product topology.
Let $f : X \times Y \to T$ be a continuous mapping.
Let $x \in X$.
Then the $x$-vertical section $f_x : Y \to T$ is continuous.
Proof
From the definition of the $x$-vertical section, we have:
- $\map {f_x} y = \map f {x, y}$
for each $y \in Y$.
Define the map $p_x : Y \to X \times Y$ by:
- $\map {p_x} y = \tuple {x, y}$
for each $y \in Y$.
We have that:
- $f_x = f \circ p_x$
From Composite of Continuous Mappings is Continuous, since $f$ is continuous, it suffices to show that $p_x$ is continuous.
From Box Topology on Finite Product Space is Product Topology, the product topology on $X \times Y$ is also the box topology.
From Continuity Test using Basis, it then suffices to check that:
Let $U$ be open in $X$ and $V$ be open in $Y$.
Note that:
- $y \in p_x^{-1} \sqbrk {U \times V}$
- $\tuple {x, y} \in U \times V$
which is equivalent to:
- $x \in U$ and $y \in V$.
Clearly then if $x \not \in U$, we have $p_x^{-1} \sqbrk {U \times V} = \O$.
We can also read off from this equivalence that if $x \in U$, we have $p_x^{-1} \sqbrk {U \times V} = V$.
From Empty Set is Element of Topology, $\O$ is open in $Y$.
Since $V$ is open in $Y$ by hypothesis, we therefore have that $p_x$ is continuous from Continuity Test using Basis.
Hence the result.
$\blacksquare$