Let $(X_j, \tau_j)_{j\in J}$ be a family of topological spaces. Let $X=\prod_{j\in J} X_j$ provided with the product topology and let $pr_k: X\to X_k$ be the projection on the $k$-th coordinate, then has $(X,\tau, (pr_j)_{j\in J})$ the following universal property:
Is $Y$ a topological space $(f_j: Y\to X_j)_{j\in J}$ a familiy of continuous functions, then there is exactly one continuous function $f:Y\to X$ such that for every $k$ holds $f_k=pr_k\circ f$
This universal property of $X$ and the function $pr_k$ characterizes $(X,\tau, (pr_j)_{j\in J})$ unique up to homeomorphism.
I have a question why this characterizes up to homeomorphism.
The proof goes as follows:
Suppose $P$ is a topological space with functions $\beta_j: P\to X_j$ which has the same universal property.
With that we get:
Now observe $\alpha\circ \beta: P\to P$ (Note that the $X$ in the following picture is supposed to be $P$)
Then $\alpha\circ\beta =\operatorname{id}_P$, because of $\alpha$ and $\beta$ beeing unique.
Similarly $\beta\circ\alpha = \operatorname{id}_X$ so $\alpha$ and $\beta$ are homeomorphic.
My question:
Why is $\alpha\circ\beta=\operatorname{id}_P$. And what has the uniqueness to do with it?
From the diagramm in the last picture we get that
$\beta_j\circ(\alpha\circ\beta)=\beta_j$ which implies $\alpha\circ\beta=\operatorname{id}_P$ immediatly.
Since $\beta_j\circ (\alpha\circ\beta(p))=\beta_j(p)\Leftrightarrow \alpha\circ\beta(p)=p$ for every $p\in P\Leftrightarrow \alpha\circ\beta=\operatorname{id}_P$.
What has the uniqueness to do with it?
Thanks in advance and excuse me for these awful images, but creating such diagrams on this website is always an odysse on its own…
Best Answer
You know that $\alpha: X \to P$ satisfies $$\forall j: \beta_j \circ \alpha = \text{pr}_j\tag{1}$$
and $\beta: P \to X$ satisfies $$\forall j : \text{pr}_j \circ \beta= \beta_j \tag{2}$$
now using $(1)$ and $(2)$ we get that for any $j$:
$$\text{pr}_j \circ (\beta \circ \alpha) = (\text{pr}_j \circ \beta) \circ \alpha = \beta_j \circ \alpha = \text{pr}_j\tag{3}$$
And in the diagram we get from applying the universal property for $X$ to the test space $X$ itself, we are promised a unique $\gamma: X \to X$ such that
$$\forall j: \text{pr}_j \circ \gamma = \text{pr}_j\tag{4}$$
Now, $(3)$ tells us that $\gamma = \beta \circ \alpha$ also obeys $(4)$ and standard common sense (or the axioms of a category, in a more abstract setting) tells us that $\gamma=\text{id}_X$ also satisfies $(4)$.
So the unicity of $\gamma$ tells us $\beta \circ \alpha = \gamma = \text{id}_X$ and in particular, $$\beta \circ \alpha = \text{id}_X$$
The $\alpha \circ \beta = \text{id}_P$ follows similarly from applying the universal property of $P$ to $P$ and a similar small computation to $(3)$.