Category theory exercise: monomorphism and fibered product How can I show that
if a morphism $\pi: X \rightarrow Y$ is a monomorphism
then the fibered product $X \times_{Y} X$ exists and the induced
morphism $X \times_{Y} X \rightarrow X$ is an isomorphism?
Thanks!
 A: You have a commutative diagram:
$$\begin{array}{ccc} X & \xrightarrow{=} & X \\ \downarrow = & & \downarrow \pi \\ X & \xrightarrow{\pi} & Y \end{array}$$
This induces a morphism $$f : X \to X \times_Y X$$ Let $$p : X \times_Y X \to X$$ be the canonical projection morphism (the two canonical projections are equal by symmetry). Then $f$ and $p$ are inverse to each other:


*

*$\pi p f = \pi = \pi \circ id_X$ by the definition in the commutative diagram I wrote; since $\pi$ is a monomorphism, $p f = id_X$.

*$f p : X \times_Y X \to X \times_Y X$ is uniquely determined by its projections on $X$ (because of the UMP of the fibred product). But $p f p = p$ by the first bullet point, and therefore $f p = id_{X \times_Y X}$.



Here's another (completely equivalent) proof: we'll show that the commutative square I wrote at the beginning satisfies the UMP of the fibred product, hence $X \cong X \times_Y X$. Suppose you got a cone over $X \xrightarrow{\pi} Y \xleftarrow{\pi} X$, ie. a square:
$$\begin{array}{ccc} Z & \xrightarrow{f} & X \\ \downarrow g & & \downarrow \pi \\ X & \xrightarrow{\pi} & Y \end{array}$$
Since $\pi$ is a monomorphism, it follows that $f = g$. So there is a unique morphism $Z \to X$ (namely $f = g$) to the first cone I defined.
