If $X\times Y\rightarrow Z$ is continuous then the partial maps $\langle\cdot , y_{0}\rangle$ and $\langle x_{0}, \cdot\rangle$ are continuous Let $f:X\times Y\rightarrow Z$ be a mapping of topological spaces, the product given the product topology.  Then for all $y_{0}\in Y$, the map $f(.,y_{0}):X\rightarrow Z$ is continuous.  Any hints on how to show this?
Thanks.
 A: Consider the maps $g_1:X\to Y$ given by $x\mapsto y_0$ and $g_2:X\to X$ given by $x\mapsto x$. They are obviously continuous, since $g_1$ is constant and $g_2$ is the identity. Now by the universal property of products, we get a unique continuous map 
$i_{y_0}:X\to X\times Y$ such that $\pi_Y\circ i=g_1$ and $\pi_X\circ i=g_2$. Here $\pi_X$ and $\pi_Y$ are of course the continuous projections of $X\times Y$ onto $X$ resp. $Y$.
This just says that $i_{y_0}:x\mapsto (x,y_0)$ is continuous.
Now as Arturo already said in the comments, we have $f(\cdot,y_0)=f\circ i_{y_0}$, which is continuous as the composition of continuous functions. 
A similar argument holds for continuity of $f(x_0,\cdot)$.
Personally I usually prefer these kind of arguments using the universal property, they feel more elegant than working explicitly with opens and inverse images. 
Alternatively, the use of nets also gives a nice proof. Use that the product topology is such that a net in a product converges iff all projections converge. Also use that a function is continuous iff it sends converging nets to converging nets. So suppose $x_i\to x$ is a net in $X$ with limit $x$. Then $(x_i,y_0)_i$ is a net converging to $(x,y_0)$. Hence, as $f$ is continuous, $f(x_i,y_0)_i$ converges to $f(x,y_0)$. In other words, $g(x_i)\to g(x)$, where $g=f(\cdot, y_0)$, which shows $g$ is continuous.
