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I'm using the definition of fibre product from the Stacks Project.

How do I show that $h_{x \times_y z}(w) = h_{x}(w) \times_{h_y(w)} h_z(w)$ ?

Is equality correct or should it be isomorphism?

I'm guessing isomorphism. If $p, q$ are the respective maps in the above article, then $-\circ p, -\circ q$ are the corresponding maps in the diagram:

$$ \array { h_{x}(w) \times_{h_y(w)} h_z(w) & \xrightarrow{-\circ q} & h_z(w) \\ \downarrow - \circ p & \ & \downarrow - \circ g \\ h_{x}(w) & \xrightarrow{-\circ f} & h_y(w) } $$

Now we also have maps $-\circ \alpha : h_{x \times_y z} (w) \to h_x(w), \ -\circ \beta : h_{x\times_y z} (w) \to h_z(w)$

Clearly since $f\circ \alpha = g \circ \beta$ in the article, then we have that $-\circ f\circ \alpha = - \circ g \circ \beta$.

But we need the reverse of this situation namely, given $\beta': h_{x \times_y z}(w)\to h_z(w)$ we need a $\beta$.

But there is an isomorphism $\gamma$ in the article such that $\gamma : w \simeq x \times_y z$.

Suppose that $-\circ g\circ \beta' = -\circ f\circ \alpha' : h_{x \times_y z}(w) \to h_y(w)$. Then let $\alpha = p\circ\gamma\circ f \circ \alpha' \circ \gamma^{-1}$.

Then there is a unique morphism $\gamma' : w \to x \times_y z $ such that $p\circ\gamma' = \alpha = p \circ \gamma \circ f \circ \alpha'$. Thus, $\gamma' = \gamma \circ f \circ \alpha'\circ \gamma^{-1}$ and $f\circ \alpha' = \gamma^{-1} \circ \gamma' \circ \gamma : w \to $

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    $\begingroup$ It should be an isomorphism. I would not be surprised in the Stacks Project has some convention about that. You should prove the general fact that $\text{Hom}(X,\text{Lim}D) \cong \text{Lim}(\text{Hom}(X,-)\circ D)$. $\endgroup$ – Derek Elkins May 29 '17 at 22:27
  • $\begingroup$ Without proving the general case, isn't this almost immediate by definition of pullback in $Set$? $\endgroup$ – Matematleta May 29 '17 at 22:31
  • $\begingroup$ @DerekElkins I'm not at limits yet. I'm going in order of the stacks project. $\endgroup$ – BananaCats Category Theory App May 29 '17 at 22:31
  • $\begingroup$ @FruitfulApproach Not specific to your question, but you may find math.meta.stackexchange.com/questions/2324/… useful. $\endgroup$ – Derek Elkins May 29 '17 at 23:08
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As mentioned by @DerekElkins in the comments, this should be an isomorphism. (In fact, it is natural.)

To see this, note that the universal property says that $h_{x\times_y z}(w) = \mathrm{Hom}_\mathcal{C}(w, x \times_y z)$ is in bijective correspondence with the set $$ \{ (\alpha, \beta) \in \mathrm{Hom}_\mathcal{C}(w, x) \times \mathrm{Hom}_\mathcal{C}(w,z) : f \alpha = g \beta \}.$$ On the other hand, this set is precisely the pullback $h_x(w) \times_{h_y(w)} h_z(w)$.

I have gently skimmed some details. In particular, I have mentioned nothing about naturality. Please let me know if you would like me to put them in.

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$h_{x}(w) \times_{h_y(w)} h_z(w)$ is a pullback of $sets$ and so is the set of pairs $(\alpha,\beta)$ such that the following square commutes:

\begin{array}{c} h_x(w) \times_{h_y(w)} h_z(w) & \overset{\pi}{\rightarrow} & h_x(w) \\ \ \pi \downarrow&& f^*\downarrow \\ h_z(w) & \overset{g^*}{\rightarrow} & h_y(w) \end{array}

Thus, $f^*\circ \pi\circ \alpha=g^*\circ \pi\circ \beta\Rightarrow f\alpha=g\beta. $

Now, consider the diagram which I copied from your link:

enter image description here

According to the definition of fibre product (pullback), there is an isomorphism

$\gamma \mapsto(\alpha,\beta)$ such that $f\alpha=g\beta;\ $ i.e., $h_{x \times_y z}(w) \cong h_{x}(w) \times_{h_y(w)} h_z(w)$

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