What is the product in the category of sets with only the injections as maps?

In the category of sets with all the functions/maps, the product will be the "$\times$", but in the subcategory with only the injections, "$\times$" doesnt fit, and the intersection doesn't appear to fit either. Take the sets $\{1, 2\}$ and $\{1,3\}$. So $\{1\}$ is the intersection, but one singleton set with the injection $1 \mapsto 3$ can't make the diagram commute with the inclusion. Maybe it will be the intersection, but with which function, and why?

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    $\begingroup$ I edited your question to include the question in the body, fix spelling and grammar, and add proper formatting. Some advice on asking a good question: math.stackexchange.com/help/how-to-ask $\endgroup$ Commented May 23 at 23:11

3 Answers 3


This category does not have binary products (or a terminal object, i.e., the empty product).

To see this, let $1$ be a set with one element, and let $2$ be a set with two elements. Suppose for contradiction that a product $1\times 2$ exists. Then since the map $\pi_1\colon 1\times 2\to 1$ is injective, $1\times 2$ has at most one element.

There are two pairs of arrows $(f,g)$ with $f\colon 1\to 1$ and $g\colon 1\to 2$, so there should be two arrows $1\to 1\times 2$. But there is at most one such arrow, since $1\times 2$ has at most one element. Contradiction.


Products of pairs of sets that aren't both singletons don't exist in your category, which I will prove by contradiction. A product of $X,Y$ would have to be a set $X \cdot Y$ equipped with injections $p_X: X \cdot Y \to X, p_Y: X\cdot Y \to Y$. I assume such a product exists.

Without loss of generality, assume that $|X| \le |Y|$. Then there exists an injection $f: X \to Y$. Then by applying the definition of the product to $f: X \to Y$ and $\operatorname{Id}_X : X \to X$ we see that there is an injection $\phi: X \to X\cdot Y$ such that $p_X \circ \phi = \operatorname{Id}_X.$ In particular, we deduce that $p_X$ is a surjection so that since it was assumed to be injective it is bijective and hence invertible.

In particular, for an arbitrary $Z$ with injections $g: Z \to X$ and $h: Z \to Y$, we must have that the unique map $Z \to X \cdot Y$ such that the diagram defining the product commutes is equal to $p_X^{-1} \circ g$. But this implies by commutativity of the diagram that $h = p_Y \circ p_X^{-1} \circ g$. Since $h$ was an arbitrary injection from $Z$ to $Y$ we conclude the proof of my claim by just exhibiting two different injections (here is where I use the assumption that $Y$ is not a singleton).

  • $\begingroup$ Why does $|X\cdot Y|=|X|$ imply that $p_X$ is a bijection? Are you assuming $X$ is finite? $\endgroup$
    – Servaes
    Commented May 24 at 19:05
  • $\begingroup$ @Servaes Good catch! At the time of writing I was (because I was converting the first counterexample that came to mind into something more general as I typed) but I don't think I needed to since that part of the argument really showed that $p_X$ has a right inverse so that the cardinality argument wasn't necessary. $\endgroup$ Commented May 24 at 20:03

Let's write $\bf Inj$ for the category of sets with injective functions.

To prove that binary products do not always exist in $\bf Inj$, recall that they must satisfy (in ${\bf Set}$): $$ {\bf Inj}(A,B\times_{{\bf Inj}} C) \cong {\bf Inj}(A,B) \times_{\bf Set} {\bf Inj}(A,C) $$ Moreover, if $1$ is a singleton, we also have (in ${\bf Set}$) $$ {\bf Inj}(1,X) \cong {\bf Set}(1,X) \cong X $$

Combining both facts we can see that the product $B\times_{{\bf Inj}} C$ must have the same cardinality as the cartesian product $B\times_{{\bf Set}} C$. Indeed, we can have (in ${\bf Set}$): $$ \begin{array}{l} B\times_{{\bf Inj}} C \\ \cong {\bf Inj}(1,B\times_{{\bf Inj}} C) \\ \cong {\bf Inj}(1,B) \times_{\bf Set} {\bf Inj}(1,C) \\ \cong B \times_{\bf Set} C \end{array} $$

To obtain a contradiction, let $2,3$ be sets with that cardinality. We get this sequence (in ${\bf Set})$: $$ \begin{array}{l} \emptyset \not\cong {\bf Inj}(3,2\times_{{\bf Set}} 2) \\ \cong {\bf Inj}(3,2\times_{{\bf Inj}} 2) \\ \cong {\bf Inj}(3,2) \times_{{\bf Set}} {\bf Inj}(3,2) \\ \cong \emptyset \times_{{\bf Set}} \emptyset \cong \emptyset \end{array} $$ Contradiction.


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