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

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?

• 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 Commented May 23 at 23:11

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).

• Why does $|X\cdot Y|=|X|$ imply that $p_X$ is a bijection? Are you assuming $X$ is finite? Commented May 24 at 19:05
• @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. 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.