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, 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:


 * $p_x^{-1} \sqbrk {U \times V}$ is open whenever $U$ is open in $X$ and $V$ is open in $Y$.

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.