Equaliser of equal arrows I have to show that, if $f,g:X\to Y$ in a category $C$ then the following are equivalent
(a) $f=g$
(b) $f$ and $g$ has some equaliser and it is an isomorphism
(c) $f$ and $g$ has some coequaliser and it is an ismomorphism.
Is my way of thinking correct?
I say that if (b) is assumed to be true then I can take $(X, id_X)$ as an equaliser and hence $(b)\Rightarrow (a)$, but then since equalisers are defined by universal property they are isomorphic and here I have some problems since this means that all the possible equalisers are of this form and I just cant see why. I mean this feels a bit wrong.  Any help?
 A: Well I would dare to say that it is obvious that $(a) \Rightarrow (b)$ and $(a) \Rightarrow (c)$: 


*

*($(a)\Rightarrow (b)$) clearly if $f=g$ then $\text{id}_X$ equalizes $f$ and $g$ and every other morphism which equalizes them factors through $\text{id}_X$, which proves that $\text{id}_X$ is the equalizer

*($(a) \Rightarrow (c)$) follows by duality from the previous point.


The other parts, while being a little less obvious, can be proven basically with your argument.


*

*($(b) \Rightarrow (a)$) if $f$ and $g$ have an equalizer which is an isomorphism this means that is must be in the form $(Z,\gamma)$ where $$\gamma \colon Z \longrightarrow X$$ is an isomorphism. Clearly also $(X,\text{id}_X)=(X,\gamma \circ \gamma^{-1})$ is an equalizer too, by abstract nonsense and so it follows that $f=g$.

*($(c) \Rightarrow (a)$) I would dare to say that it follows by duality too.
Now back your doubt. While is it true that all the equalizers of two morphisms, say $f$ and $g$, are isomorphic this does not mean that they are the same. 
Remember that an equalizer is a pair $(Z,\gamma)$ formed by an object and a morphism and it is isomorphic to another equalizer $(Z',\gamma')$ if and only if there is an $\alpha \colon Z \to Z'$ such that $\gamma=\gamma'\circ\alpha$ and $\gamma'=\gamma\circ\alpha^{-1}$. 
Clearly this $\alpha$ is not required to be an identity so there is no reason why if $\gamma=\text{id}_X$ then $\gamma'=\text{id}_{Z'}$ too.
Hope this helps.
A: If $(E,e)$ is an equalizer of two arrows $f:X\to Y$, $g:X\to Y$ and $\varphi:E'\to E$ is an isomorphism, then $(E',e\varphi)$ is also an equalizer. Indeed, if $ft=gt$ for some arrow $t:Z\to X$, then $t=eu$ for some arrow $u:Z\to E$ and thus $t=e\varphi(\varphi^{-1}u)$ factorises through $e\varphi$; and such a factorization has to be unique since $e\varphi$ is a composition of monomorphisms. In particular, if $e$ is an isomorphism with inverse $e^{-1}:X\to E$, then $(X,id_X)=(X,ee^{-1})$ is an equalizer for $f$ and $g$. And then any isomorphism $\varphi:E'\to E$ is also an equalizer.
Conversely, if $(E',e')$ is an equalizer for $f,g$ then there exists a unique isomorphism $\varphi:E'\to E$ such that $e'=e\varphi$. Thus if one of the equalizers is an isomorphism, all of them must be isomorphisms as well.
A: If $f,g:X\to Y$ in a category $C$ then the following are equivalent
(a) $f=g$;
(b) $f$ and $g$ has some equaliser and it is epic;
(c) $f$ and $g$ has some coequaliser and it is monic;
$(b\Rightarrow a)$
If $e:\bullet\to X$ is an equalizer for $f$ and $g$ then $ef=eg$; if $e$ is epic then this implies $f=g$.
$(c\Rightarrow a)$
Dual of $(b\Rightarrow a)$.
$(a\Rightarrow b)$
If $f=g$ then $1_X:X\to X$ equalizes $f$ and $g$ and this is epic.
$(a\Rightarrow c)$
Dual of $(a\Rightarrow b)$
