Consider the pointed category $\mathbf B$ (category where each morphism classes hom$(A,B)$ has a zero morphism). Then I want to prove the fact that this category has to be a full subcategory of a category $\mathbf A$ with zero object.

But why it is always the case since we can remove some morphisms from $\mathbf A$ without breaking the feature to be pointed category?

So what has to be the way of evidence? Give some help if you can.

  • 1
    $\begingroup$ Note that in general pointed category means "category with a zero object" rather than "category enriched over pointed sets". $\endgroup$ – Arnaud D. Nov 20 '17 at 9:47

Given a category $B$ with zero morphisms $0_{XY}:X\to Y$ between any two objects, you can obtain a category $A$ with a zero object by simply adding a zero object to $B$. More precisely, this means that :

  • $Ob(A)=Ob(B)\sqcup \{0\}$
  • if $X,Y\in Ob(B)$, then $Hom_A(X,Y)=Hom_B(X,Y)$
  • if $X\in Ob(B)$, then $Hom_A(0,X)=\{\alpha_X\}$ and $Hom_A(X,0)=\{\tau_X\}$
  • $Hom_A(0,0)=\{id_0\}$
  • composition of morphisms of $B$ defined as in $B$
  • for every $X,Y\in Ob(B)$, $\tau_X\circ \alpha_X=id_0$ and $\alpha_Y\circ\tau_X=0_{XY}$
  • for every $f:X\to Y$ in $B$, $f\circ \alpha_X=\alpha_Y$ and $\tau_Y\circ f=\tau_X$

You can check that all this defines a category, and that $0$ is a zero object of $A$. Moreover, $B$ is a subcategory of $A$, and it is full thanks to the second point.

Note also that if there was already a zero object $Z$ in $B$, then $0_{ZZ}=id_Z$; together with the sixth point, this implies that $Z\cong 0$ in $A$. In that case $A$ and $B$ are equivalent.

  • $\begingroup$ but If I'll add more morphisms to the morphism classes of a category B? $\endgroup$ – A. Gonus Nov 23 '17 at 13:13
  • $\begingroup$ I'm not sure I understand your comment. But if you look at my second bullet point, you see that I do not add any morphism between objects in $B$; I only add a new object (first bullet point) and morphisms to and from this new object (third bullet point). This is why $B$ is a full subcategory of $A$. $\endgroup$ – Arnaud D. Nov 23 '17 at 14:21
  • $\begingroup$ I think, I fully understand Your answer, but I’m also interested if it is always the case, that is, can we obtain a category A by not only adding a zero object to B, but also adding more morphisms to its hom-sets? $\endgroup$ – A. Gonus Nov 23 '17 at 21:14
  • $\begingroup$ If you add more morphisms $B$ will not be a full subcategory... $\endgroup$ – Arnaud D. Nov 23 '17 at 21:20
  • $\begingroup$ yes, I understand this, so it is possible to do this construction and we just will lose the feature to be the full subcategory? $\endgroup$ – A. Gonus Nov 23 '17 at 21:34

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