If $F$ is a functor such that $F(eq(f,g)) = eq(F(f), F(g))$ and $F(f)$ is an isomophism, then $F$ is faithful

Question

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.

Answer 1

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 $x = y$.

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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}

Answer 2

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.