About certain regular epimorphisms in a Grothendieck Topos

Question

I am supposed to prove a rather technical property which should hold in any Grothendieck Topos, but I have troubles in accomplishing this task. Here is the context for the question.

Let then $\mathscr{E}$ be a Grothendieck Topos and suppose to have spans

$$ Y_{1}\xleftarrow{h_{1}} Y_{0}\xrightarrow{h_{2}} Y_{2},\quad X_{1}\xleftarrow{k_{1}} X_{0}\xrightarrow{k_{2}} X_{2} $$

in $\mathscr{E}$. Assume given a morphism between those spans, i.e. a triple

$$ (Y_{1}\xrightarrow{f_{1}} X_{1},\ Y_{0}\xrightarrow{f_{0}} X_{0},\ Y_{2}\xrightarrow{f_{2}} X_{2}) $$

making the following diagram commutative

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\la}[1]{\kern-1.5ex\xleftarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\las}[1]{\kern-1.5ex\xleftarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} Y_{1} & \la{h_1} & Y_{0} & \ra{h_2} & Y_{2} \\ \da{f_1} & & \da{f_0} & & \da{f_2} \\ X_{1} & \las{k_1} & X_{0} & \ras{k_2} & X_{2} \\ \end{array} \qquad\qquad (\dagger) $$ Suppose, in addition, that both the squares in $(\dagger)$ above are pullback squares, i.e. $Y_{0}=X_{0}\times_{X_{1}} Y_{1}$ and $Y_{0}=X_{0}\times_{X_{2}} Y_{2}$.

Now, I can take the pushout of the first and of the second row of $(\dagger)$, so as to get pushout diagrams

$$ \begin{array}{c} Y_{0} & \ra{h_2} & Y_{2} \\ \da{h_1} & & \da{\pi_{Y_{2}}} \\ Y_{1} & \ras{\pi_{Y_{1}}} & Y\\ \end{array} \quad\quad \begin{array}{c} X_{0} & \ra{k_2} & X_{2} \\ \da{k_1} & & \da{\pi_{X_{2}}} \\ X_{1} & \ras{\pi_{X_{1}}} & X\\ \end{array} $$

By functoriality of the pushout, we then get an induced map $f\colon Y\rightarrow X$ which is such that $f\circ \pi_{Y_{j}}=\pi_{X_{j}}\circ f_{j}$, for $j\in\{1,2\}$. I can then take the pullbacks (again for $j\in\{1,2\}$)

$$ \begin{array}{c} X_{0}\times_{X} Y & \ra{}& Y \\ \da{} & & \da{f} \\ X_{0} & \ras{\pi_{X_{1}}\circ\ k_{1}} & X\\ \end{array} \quad\quad \begin{array}{c} X_{j}\times_{X} Y & \ra{}& Y \\ \da{} & & \da{f} \\ X_{j} & \ras{\pi_{X_{j}}} & X\\ \end{array} \quad\quad (\ddagger) $$

Finally, I have pairs of maps $(Y_{0}\xrightarrow{f_{0}} X_{0},\ Y_{0}\xrightarrow{\pi_{Y_{1}}\circ\ h_{1}} Y)$ and $(Y_{j}\xrightarrow{f_{j}} X_{j},\ Y_{j}\xrightarrow{\pi_{Y_{j}}} Y)$, for $j\in\{1,2\}$, which are compatible with the pullbacking maps in $(\ddagger)$, so that I get maps $g_{l}\colon Y_{l}\rightarrow X_{l}\times_{X} Y$ (here $l\in\{0,1,2\}$) satisfying the prescribed commutativity properties.

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Here is the question: I need to prove that each such $g_{l}$ is a regular epimorphism.

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Actually, I would already be satisfied if I could prove the claim for $\mathscr{E}=\mathbf{Set}$, hopefully using some categorical arguments (i.e. topos-theoretic properties of $\mathbf{Set}$), but even an elementwise proof may be good. I am completely stuck with the task of showing this result: I have tried to write down explicitely the maps $g_{l}$ but I am unable to prove their surjectivity. I have also tried to write down some more diagrams, muttering something about the effectiveness of equivalence relations in $\mathbf{Set}$ (actually, in any Grothendieck Topos) and the fact that the pullback functors along a map $g\colon A\rightarrow B$ in $\mathbf{Set}$ (actually, in any (Grothendieck) Topos) preserve colimits between the appropriate slice categories. The idea behind this attempt was that of proving that each $g_{l}$ is the coequaliser of its kernel pair (which must be true, if the result has to hold), but I could not show this fact.

Any help (solutions or ideas) would be greately appreciated.

Answer 1

As discussed, it suffices to prove the claim for $\mathcal{E} = \mathbf{Set}$.

Let $x_1 \in X_1$, $y \in Y$, and suppose $\pi (x_1) = f (y)$. Consider $y$:

• Either $y = \pi (y_1)$ for some $y_1 \in Y_1$,
• or $y = \pi (y_2)$ for some $y_2 \in Y_2$, in which case $\pi (x_1) = f (\pi (y_2)) = \pi (f_2 (y_2))$; but that implies $f_2 (y_2) = k_2 (x_0)$ for some $x_0 \in X_0$, so there is a unique $y_0 \in Y_0$ such that $f_0 (y_0) = x_0$ and $h_2 (y_0) = y_2$, and thus for $y_1 = h_1 (y_0)$, we have $y = \pi (y_1)$.

Thus we may assume $y = \pi (y_1)$ for some $y_1 \in Y_1$. The trick is in choosing $y_1$ so that $f_1 (y_1) = x_1$. For this, observe that there is a unique function $d : X_1 \times X_1 \to \mathbb{N} \cup \{ \infty \}$ satisfying the following conditions:

• We have $a = b$ if and only if $d (a, a) = 0$.
• Given elements $r_1, s_1, \ldots, r_n, s_n$ in $X_0$ such that \begin{align} f_2 (r_k) & = f_2 (s_k) && 1 \le k \le n \\ f_1 (r_{k+1}) & = f_1 (s_k) && 1 \le k < n \end{align} then $d (f_1 (r_1), f_1 (s_n)) \le n$.
• If $d (a, b) = n$ and $0 < n < \infty$, then there exist elements $r_1, s_1, \ldots, r_n, s_n$ in $X_0$ such that \begin{align} f_2 (r_k) & = f_2 (s_k) && 1 \le k \le n \\ f_1 (r_{k+1}) & = f_1 (s_k) && 1 \le k < n \end{align} and $f_1 (r_1) = a$ and $f_1 (s_n) = b$.

It follows from these conditions that $d$ is a metric on $X_1$. Now:

• If $f_1 (y_1) = x_1$, then we are done.
• If $f_1 (y_1) \ne x_1$, then $0 < d (f_1 (y_1), x_1) < \infty$, so for some $r$ and $s$ in $X_0$, we have $k_1 (r) = f_1 (y_1)$, $k_2 (r) = k_2 (s)$, and $d (k_1 (s), x_1) < d (f_1 (y_1), x_1)$. But then there is a unique $r' \in Y_0$ such that $f_0 (r') = r$ and $h_1 (r) = y_1$, so $f_2 (h_2 (r')) = k_2 (f_0 (r')) = k_2 (r) = k_2 (s)$, hence there is a unique $s' \in Y_0$ such that $f_0 (s') = s$ and $h_2 (s') = h_2 (r')$, and therefore $f_1 (h_1 (s')) = k_1 (f_0 (s')) = k_1 (s)$.

Thus, by induction on $d (f_1 (y_1), x_1)$, we deduce there is always some choice of $y_1 \in Y_1$ such that $f_1 (y_1) = x_1$. This proves that $Y_1 \to X_1 \times_X Y$ is surjective. By symmetry, $Y_2 \to X_2 \times_X Y$ is also surjective.

It remains to be shown that $Y_0 \to X_0 \times_X Y$ is surjective. Let $x_0 \in X_0$, $y \in Y$, and suppose $\pi (k_1 (x_0)) = f (y)$. By the above, there is $y_1 \in Y_1$ such that $f_1 (y_1) = k_1 (x_0)$ and $\pi (y_1) = y$. Then there is a unique $y_0 \in Y_0$ such that $h_2 (y_0) = y_1$ and $f_0 (y_0) = x_0$. This completes the proof.

Answer 2

EDIT Okay I think I at least half-believe the statement should be true now. Think of the case where the maps in the pushout diagram $X = X_1 \cup_{X_0} X_2$ are all injections so that $X_1,X_2$ are just two subsets of $X$ and $X_0$ is their intersection. Think of the $Y_i \to X_i$ as bundles. Then $Y \to X$ is the bundle you get by gluing these together, and $X_i \times_X Y \to X_i$ is the restriction of the glued bundle back to the gluing piece. What you want to show is that the restriction of the glued bundle is just the bundle you started with, which seems intuitive. I think this is related to the Beck-Chevalley condition.

[EDIT The following was an attempted counterexample which was not applicable to the setup as pointed out in the comments.]

I'm not sure I've followed every twist of the construction, but take the $X_i$ to all be the one-point set. Then $X$ is also the one-point set. So $X_i \times_X Y = Y$, and you have maps $Y_i \to Y$ that you want to be surjective. But it's easy to arrange that some of the $Y_i$ have smaller cardinality than $Y$ (for instance if $Y_i$ is the $i$-point set :), so no such surjection exists.

Intuitively, I wouldn't expect this statement to be true because although there are lots of ways to preserve epimorphisms, there are not so many ways (that I'm aware of) to construct epimorphisms. So it's kind of fishy there's no hypothesis that something starts out epimorphic.