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Pullbacks exists but are not what you describe.

The answer is as follows:

The category $\mathcal{C}$ of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set $A$ to $A$ disjoint union a base point $*$ and sending $f$ to the obvious based function which sends everything on which $f$ was not defined to the base-point.

The pullback in based sets are well-known and for example the product $\times$ translating in based sets translates back through this equivalence to the category of sets $\mathcal{C}$ and partial functions becomes:

$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B$

From the purely sets and partial function point of view this is also explainable: . Indeed, any morphism from $Z$ to this product is given by a choice for each point in $Z$ of either: a point in $A$ and a point in $B$, or a point in $A$, or a point $B$.B$, or nothing.

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Pullbacks exists but are not what you describe.

The answer is as follows:

The category $\mathcal{C}$ of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set $A$ to $A$ disjoint union a base point $*$ and sending $f$ to the obvious based function which sends everything on which $f$ was not defined to the base-point.

The pullback in based sets are well-known and for example the product $\times$ but translating back through this equivalence to the category of sets and partial functions you essentially getbecomes:

$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B$

From the purely sets and partial function point of view this is also explainable: any morphism from $Z$ to this product is given by either a point in $A$ and a point in $B$, or a point in $A$, or a point $B$.

show/hide this revision's text 1

Pullbacks exists but are not what you describe.

The answer is as follows:

The category $\mathcal{C}$ of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set $A$ to $A$ disjoint union a base point $*$ and sending $f$ to the obvious based function which sends everything on which $f$ was not defined to the base-point.

The pullback in based sets are $\times$ but translating back through this equivalence to the category of sets and partial functions you essentially get:

$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B$