TITLE: Sophism: Discrete topology is a product topology
QUESTION [2 upvotes]: Let $X_1$, $X_2$ be topological spaces. Let $f_1:Y\rightarrow X_1$ and $f_2:Y\rightarrow X_2$. I will construct a (fake) product $X_1\times X_2$.
Let $\pi_1$, $\pi_2$ be cartesian projections.
Let $f$ be the function defined by the equality $f(x,y)=(f_1(x),f_2(y))$.
Take $X_1\times X_2$ the discrete topology on $X_1\times X_2$.
For every object $Y$ and pair of morphisms $f_1: Y \to X_1$, $f_2: Y \to X_2$ there exists a unique function $f: Y \to X$ such that $\pi_1\circ f=f_1$ and $\pi_2\circ f=f_2$ ($f$ is morphism because every function from a discrete space is a morphism).
So discrete topology is a product.
Where is my error?

REPLY [3 votes]: A product of two objects $X_1,\, X_2$ in a category is an object $X$ together with a pair of morphisms $\pi_i \colon X \to X_i$ such that for every object $Y$ and pair of morphisms $f_i \colon Y \to X_i$, there exists a unique morphism $f \colon Y \to X$ such that $f_i = \pi_i \circ f$.
In the category $\mathbf{Top}$, that means continuous maps.
Now, if you endow the cartesian product (the product in the category $\mathbf{Ens}$ of sets) $X_1\times X_2$ with the discrete topology, the function $f\colon x \mapsto (f_1(x), f_2(x))$ is the unique function with $\pi_i \circ f = f_i$, but it is in general not a morphism in $\mathbf{Top}$, i.e., it is in general not continuous (unless either one of the spaces is empty, or both spaces are discrete).
For example, if $X_1$ is not discrete, and $X_2$ not empty, we have the continuous maps $f_1 = \operatorname{id} \colon X_1 \to X_1$ and $f_2 \colon X_1 \to X_2;\; x \mapsto c$ for some $c \in X_2$. Then the map $x \mapsto (x,c)$ into the cartesian product $X$ with the discrete topology is not continuous.
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