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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!

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Well, if $X \times_Y X \to X$ is an isomorphism, you may as well take $X \times_Y X = X$. Show that this works. –  Zhen Lin Mar 28 at 16:16

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up vote 1 down vote accepted

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.

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