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Let $\mathbb{A}$ be a category, let $I \in \mathbb{A}$ be the initial object of the category, then for every $A \in \mathbb{A}$ there $\exists ! f:I \longrightarrow A$.

So, i'm trying to prove that $f$ is in fact a monomorphism, i.e., given $g,h:B \longrightarrow I \in \mathbb{A}$ such that $f\circ g=f\circ h$ then $g=h$.

I can't prove it and also i can't refute it yet. Do you guys have any hint to get into the result?

Thanks.

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4  
It is often a good idea to look at examples first before trying to prove something. It already fails for the category of rings, $\mathbb{Z}$ is initial and $\mathbb{Z} \to 0$ is not a monomorphism. –  Martin Brandenburg Sep 4 '13 at 7:44
    
@MartinBrandenburg Why is $\mathbb{Z}$ initial? There are two ring homomorphisms from $\mathbb{Z}\to\mathbb{Z}$, the identity and the 0. –  mez Sep 4 at 18:00
    
Rings are unital (for me and most other people), so homomorphisms of rings are unital by definition. –  Martin Brandenburg Sep 5 at 0:23
    
@MartinBrandenburg Ok, but then many would exclude $0$ from the category of unital rings. Some authors require at least 2 elements. –  mez Sep 5 at 0:50
    
The zero ring $0$ is an unital ring. I guess that 99% of the authors agree that $0 \neq 1$ should not be assumed in the definition of a ring, for example because otherwise you would have to make many case distinctions in even basic constructions of rings. And after all, the zero ring is just a very natural object. Likewise, the empty space is a very natural topological space, and $\mathrm{Spec}(0) = \emptyset$. –  Martin Brandenburg Sep 5 at 0:55

4 Answers 4

up vote 12 down vote accepted

Let $\mathcal{C}$ be a category whose objects are $\mathrm{ob}(\mathcal{C})=\{x,y\}$ and whose morphisms are $$\begin{align} \mathrm{Mor}_{\mathcal{C}}(x,x)&=\{\mathrm{id}_x\} & \mathrm{Mor}_{\mathcal{C}}(x,y)&=\{f\}\\\\ \mathrm{Mor}_{\mathcal{C}}(y,x)&=\{g,h\} & \mathrm{Mor}_{\mathcal{C}}(y,y)&=\{\mathrm{id}_y,k\} \end{align}$$ where $g\circ f=h\circ f=\mathrm{id}_x$ and $f\circ g=f\circ h=k$, $\;k\circ k=k$, $\;g\circ k=g$, $\; h\circ k=h$.

Then $x$ is an initial object of $\mathcal{C}$ and $f\circ g=f\circ h$, but $g\neq h$.

(As Jim points out in a comment below his answer, the category I had used was not actually a counterexample; we need to include a non-identity morphism from $y$ to itself.)


Though you've already accepted my answer, I had been in the process of making a diagram of this category, and I might as well include it at this point:

                                       enter image description here

\documentclass{standalone}
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
x \ar[bend left=40]{r}{f}
\ar[loop left,out=220,in=140,distance=1cm]{}{\mathrm{id}_x}
& y \ar[bend left=40]{l}[swap]{g} \ar[bend left=70]{l}[pos=0.47]{h}
\ar[loop right,out=322,in=38,distance=1cm]{}{\mathrm{id}_y}
\ar[loop right,out=295,in=65,distance=2.5cm]{}[swap]{k}
\end{tikzcd}
\end{document}

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6  
I would like to add to this, that the way you can construct such an example is just by drawing the smallest possible diagram that satisfies the hypotheses, filling in the required compositions and identities. You then either have the proof you want or you have a counterexample that refutes it; in this case the latter. –  MJD Sep 4 '13 at 6:32
3  
Note that the rule $k \circ k = k$ is essential, as the other choice for this composition in this minimal category counterexample ($k \circ k = \mathrm{id}_y$) results in $$\mathrm{id}_y = f \circ g \circ f \circ g = f \circ \mathrm{id}_x \circ g = f \circ g$$ and therefore, easily, $g = h$. It is not trivial to produce counterexamples by this method unless you take care to check all the category and hypothesis consequences. –  Steve Powell Sep 4 '13 at 14:03

This statement is true if and only if its dual is true:

In any category, arrows to terminal objects are epimorphisms

We have an easy counter-example in $\mathbf{Set}$ to this claim: $\emptyset \to \{ \emptyset \} $

We can find a counter-example to the original statement in $\mathbf{Ring}$: $\mathbb{Z}$ is initial, but the maps $\mathbb{Z} \to \mathbb{Z} / p \mathbb{Z}$ are not monomorphisms.


What we do have, however, is that arrows from a terminal object are monomorphisms, and similarly arrows to an initial object are epimorphisms.

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2  
This exemplifies nicely how any time you’re trying to prove/disprove a general statement about categories, one of the first things to try is considering the dual statement. The “handedness” in how we think of categories means that very often, the dual may be more intuitive than the original, or have more evident counterexamples. –  Peter LeFanu Lumsdaine Sep 4 '13 at 16:34
    
Why is $\mathbb{Z}$ initial? There are two ring homomorphisms from $\mathbb{Z}\to\mathbb{Z}$, the identity and the 0. –  mez Sep 4 at 18:03
    
@mez: The zero map isn't a ring homomorphism because it fails to preserve the ring structure: in particular, $f(1) \neq 1$. In more detail, there are two conflicting conventions regarding what the word "ring" means; the convention you know doesn't have '1' as part of the ring structure, but the one I have used does include '1' as part of the ring structure. The latter convention is the only one I have ever seen Ring mean. Although admittedly, the only exposure I've ever had to the former convention is mention that it exists, and the observation that apparently some people are taught it. –  Hurkyl Sep 4 at 18:11

This is false. I don't have a "natural" example for you, maybe someone else can come up with one. But here is a category for which this fails, there are exactly two objects, $\{A, I\}$, and there are exactly $6$ morphisms:

  • $f\colon A \to I$
  • $g\colon A \to I$
  • $a\colon I \to A$
  • $\mathrm{id}_A$
  • $\mathrm{id}_I = fa = ga$
  • $af = ag$

Composition is given according to the rules above. Then $I$ is initial but $a$ is not a monomorphism because $f \neq g$ but $af = ag$.

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1  
@Lix: Your suggested edit, to make $af = ag = \mathrm{id}_A$, was rejected. The reason is because then the category so described would not be a counter example to the statement! This is because if we take $af = \mathrm{id}_A$ and compose on the left with $g$ we get $gaf = g$. But $ga = \mathrm{id}_I$ so this would mean $f = g$! –  Jim Sep 4 '13 at 8:00

It is often a good idea to look at examples first before trying to prove something. It already fails for the category of unital rings, here $\mathbb{Z}$ is initial and $\mathbb{Z} \to R$ is only a monomorphism when $R \neq 0$ and the characteristic of $R$ is zero. More generally, we have the following result:

Let $I$ be an initial object of a category. For the statements

  1. $I$ is strict initial, i.e. every morphism $X \to I$ is an isomorphism.
  2. All morphisms $X \to I$ are equal.
  3. Every morphism $I \to Y$ is a monomorphism.

we have 1. => 2. => 3.

Proof: 1. => 2. Let $f,g : X \to I$ be morphisms. By assumption $f$ is an isomorphism, so we may look at $g f^{-1}$. This is an endomorphism of $I$, hence the identity. 2 => 3. is trivial.

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