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Unless I’m misunderstanding something, the claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear to me. To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times_Z Y$ with $\pi_1[p] = [x]$, $\pi_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.)

Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x_1 \colon X_1 \to X$, $x_2 \colon X_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, take a large enough free cover of the image, and show that it factors through each $x_i$.) So pseudo-elements of $X$ correspond to subobjects/subgroups of $X$. Borceux notes this fact in the closing discussion of §1.9.

But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same. In terms of pseudo-elements, the maps $s_1, s_2 : \Q \to \Q \times \Q$ given by $s_i(x)=(x,ix)$ are not equal as pseudo-elements of $\Q \times \Q$, but their images under each projection are equal as pseudo-elements of $\Q$.

The claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear, but seems wrong. To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times_Z Y$ with $\pi_1[p] = [x]$, $\pi_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.)

Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x_1 \colon X_1 \to X$, $x_2 \colon X_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, note that in this case the projections $X_1 \times_X X_2 \to X_i$ are epi.) So pseudo-elements of $X$ correspond to subobjects/subgroups of $X$. Borceux notes this fact in the closing discussion of §1.9.

But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same. In terms of pseudo-elements, the maps $s_1, s_2 : \Q \to \Q \times \Q$ given by $s_i(x)=(x,ix)$ are not equal as pseudo-elements of $\Q \times \Q$, but their images under each projection are equal as pseudo-elements of $\Q$.

Unless I’m misunderstanding something, the claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear to me. To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times_Z Y$ with $\pi_1[p] = [x]$, $\pi_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.)

Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x_1 \colon X_1 \to X$, $x_2 \colon X_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, take a projective cover of the image, and note that it factors through each $x_i$.) So pseudo-elements of $X$ correspond to subobjects/subgroups of $X$. Borceux notes this fact in the closing discussion of §1.9.

But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same. In terms of pseudo-elements, the maps $s_1, s_2 : \Q \to \Q \times \Q$ given by $s_i(x)=(x,ix)$ are not equal as pseudo-elements of $\Q \times \Q$, but their images under each projection are equal as pseudo-elements of $\Q$.

Unless I’m misunderstanding something, the claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear to me. To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times_Z Y$ with $\pi_1[p] = [x]$, $\pi_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.)

Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x_1 \colon X_1 \to X$, $x_2 \colon X_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, take a large enough free cover of the image, and show that it factors through each $x_i$.) So pseudo-elements of $X$ correspond to subobjects/subgroups of $X$. Borceux notes this fact in the closing discussion of §1.9.

But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same. In terms of pseudo-elements, the maps $s_1, s_2 : \Q \to \Q \times \Q$ given by $s_i(x)=(x,ix)$ are not equal as pseudo-elements of $\Q \times \Q$, but their images under each projection are equal as pseudo-elements of $\Q$.

Unless I’m misunderstanding something, the claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear to me. To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times_Z Y$ with $\pi_1[p] = [x]$, $\pi_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.)

Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x_1 \colon X_1 \to X$, $x_2 \colon X_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, take a projective cover of the image, and note that it factors through each $x_i$.) So pseudo-elements of $X$ correspond to subobjects/subgroups of $X$. Borceux notes this fact in the closing discussion of §1.9.

But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same.

Unless I’m misunderstanding something, the claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear to me. To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times_Z Y$ with $\pi_1[p] = [x]$, $\pi_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.)

Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x_1 \colon X_1 \to X$, $x_2 \colon X_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, take a projective cover of the image, and note that it factors through each $x_i$.) So pseudo-elements of $X$ correspond to subobjects/subgroups of $X$. Borceux notes this fact in the closing discussion of §1.9.

But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same. In terms of pseudo-elements, the maps $s_1, s_2 : \Q \to \Q \times \Q$ given by $s_i(x)=(x,ix)$ are not equal as pseudo-elements of $\Q \times \Q$, but their images under each projection are equal as pseudo-elements of $\Q$.