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Suppose that $\mathcal C$ has equalizers of every pairs of morphisms. If $F: \mathcal C \to \mathcal D$ is a functor such that $F(eq(f,g)) = eq(F(f), F(g))$ and $F(f)$ is an isomophism, then $F$ is faithful.

I am not sure how to approach this question. If $F(f)=F(g)$ then $eq(F(f),F(f)) = F(eq(f,g))$. It seems to me that I must use the fact that $F(f)$ is an isomorphism to "get rid of" the $F$, but I am not sure how to do this, much less gettting a graphical intuition of this.

Any help would be greatly apporeciated.

Squirrel-Power
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    I think you're missing some quantifiers here. I guess you want $F(eq(f,g)) = eq(F(f), F(g))$ to hold for all parallel pairs $f,g$, but what about $F(f)$ being an isomorphism? Do you want that $F$ maps every arrow to an iso? – Arnaud D. Oct 12 '18 at 11:19
  • In fact, I suspect that you are asking for something like this : https://math.stackexchange.com/questions/2730781/when-does-conservative-imply-faithful-for-functors – Arnaud D. Oct 12 '18 at 11:45

2 Answers2

4

You may mean this:

Let $F:\scr C\to\scr D$ be a functor between categories. If $\scr C$ has equalizers, $F$ preserves them and reflects isomorphisms, then $F$ is faithful.

Let $f,g:X\to Y$ be a pair of parallel arrows in $\scr C$ and assume $F(f)=F(g)$. Then the equalizer of $F(f)$ and $F(g)$ is an isomorphism. Since $F$ preserves equalizers and reflects isomorphisms, it follows that the equalizer of $f$ and $g$ is an isomorphisms as well. Consequently $f=g$, hence $F$ is faithful.

Fabio Lucchini
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Fleshing out Fabio's solution:

  1. The “equaliser” of $x$ and $y$ is the best $e$ with $x ∘ e = y ∘ e$, in that any other arrow, $d$, with this property uniquely factors through it, $d = e ∘ u$.

    Intuitively the equaliser of $f, g$ is the largest sub-part of their source on which they are identified.

  2. If $x = y$ then their equaliser is the identity, which is an isomorphism.

    Intuitively the largest sub-part of $f$'s source that identifies it with itself is just the whole source.

  3. If the equaliser of $x$ and $y$ is an isomorphism, then they are equal.

       x = y
⇐  x ∘ Id = y ∘ Id
⇐  x ∘ e ∘ e⁻¹ = y ∘ e ∘ e ⁻¹
⇐  x ∘ e = y ∘ e
⇐ true, since e equalises x and y

  4. Hence, by 2 & 3, the equaliser of $x$ and $y$ is an isomorphism precisely when $x = y$.

Now we show

If $F$ preserves equalisers, where the source category has all equalisers, and reflects isomorphisms, then it is faithful; i.e. injective on arrows.

Since the source category has all equalisers, we can let $e$ be the equaliser of $f$ and $g$, we now prove $F f = F g \implies f = g$ as follows:

\begin{align} & F f = F g \\ \iff & F e \text{ iso } & \text{[$F$ preserves equalisers and (4)]} \\ \implies & e \text{ iso } & \text{[$F$ reflects isomorphisms]} \\ \iff & f = g & \text{[Applying (4) one last time ]} \end{align}

Musa Al-hassy
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