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Set theoretically, we have that for any three sets, it holds that

$$X \subseteq Y \implies X \times Z \subseteq Y \times Z.$$

Categorially, we have that if $X$ and $Y$ are objects of a category and $f : X \rightarrow Y,$ then if the products $X \times Z$ and $Y \times Z$ exist, then there is a unique arrow $g : X \times Z \rightarrow Y \times Z$ making the relevant diagram commute.

This is probably a silly question, but if $f$ is monic, does it necessarily follow that $g$ is monic?

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3 Answers 3

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The diagram

$\begin{array}{c} X \times Z & \rightarrow & Y \times Z \\ \downarrow && \downarrow \\ X & \rightarrow & Y \end{array}$

is a pullback (see here). But pullbacks of monics are monic, by a simple diagram chase; just look long enough at the following diagram:

$\begin{array}{c} T & \rightrightarrows & X' & \rightarrow & Y' \\ &&\downarrow && \downarrow \\ && X & \hookrightarrow & Y \end{array}$

This also appears somewhere in Mac Lane's book.

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  • $\begingroup$ I like this! Maximal simplicity. $\endgroup$ Aug 31, 2013 at 10:18
  • $\begingroup$ Almost. Zhen has topped it. :) $\endgroup$ Aug 31, 2013 at 10:26
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Denote $p_X,p_Z$ the projections $X \times Z \to X, Z$ and $q_Y, q_Z$ the projections $Y \times Z \to Y,Z$. From the maps $ p_X f : X \times Z \to Y$ and $p_Z : X \times Z \to Z$ we get the induced map $g : X \times Z \to Y \times Z$, such that $p_X f = g q_Y$ and $p_Z = g q_Z$.

Suppose we have an arrow $h : W \to X \times Z$. $h$ is uniquely determined by its components $hp_X$ and $hp_Z$. Since $f$ is a monomorphism, $hp_X$ is uniquely determined by $hp_Xf$, so that $h$ is determined by $hp_Xf = hgq_Y$ and $hp_Z = hgq_Z$. Hence $h$ is determined by $hg$, which means that $g$ is also a monomorphism.


you can also argue that, under the identification $\hom(W,A \times B) \cong \hom(W,A) \times \hom(W,B)$, $g_* = f_* \times id_{\hom (W,Z)}$. Since $f_*$ is injective, so is $g_*$.

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  • $\begingroup$ You seem to use the postfix notation for composition? $\endgroup$ Aug 31, 2013 at 10:09
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The most conceptual answer is that the functor ${-} \times {-} : \mathcal{C} \times \mathcal{C} \to \mathcal{C}$ is a right adjoint, at least when $\mathcal{C}$ has all products. In particular it preserves monomorphisms, and it is easy to see that monomorphisms in $\mathcal{C} \times \mathcal{C}$ are precisely the componentwise monomorphisms.

When $\mathcal{C}$ does not have all products, this argument can still be made to work by embedding $\mathcal{C}$ in $[\mathcal{C}^\mathrm{op}, \mathbf{Set}]$: this preserves and reflects all limits, so in particular, products and monomorphisms.

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  • $\begingroup$ The diagonal embedding $\Delta : \mathcal{C} \to \mathcal{C} \times \mathcal{C}$ is the left adjoint. More generally, for any diagram $\mathcal{J}$, ${\varprojlim}_\mathcal{J} : [\mathcal{J}, \mathcal{C}] \to \mathcal{C}$ is the right adjoint of the embedding-of-constants $\Delta : \mathcal{C} \to [\mathcal{J}, \mathcal{C}]$. $\endgroup$
    – Zhen Lin
    Aug 31, 2013 at 10:25
  • $\begingroup$ ${-} \times Z$ is also a right adjoint; its left adjoint is the forgetful functor $\mathcal{C}_{/ Z} \to \mathcal{C}$. However it requires a little more effort to see that monomorphisms in $\mathcal{C}_{/ Z}$ are the same as in $\mathcal{C}$. $\endgroup$
    – Zhen Lin
    Aug 31, 2013 at 10:27

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