Left adjoints preserve initial objects $\implies$ right adjoints preserve terminal objects

I can prove directly that right adjoints preserve terminal objects. But how to deduce this from the fact that left adjoints preserve initial objects?

I suppose one should pass to the opposite categories somehow (and left and right adjoints must be connected by some kind of duality), but I don't understand the details.

So why is left adjoint the dual of right adjoint? Well, suppose we have functors $$F: \mathcal{C} \to \mathcal{D}$$ and $$G: \mathcal{D} \to \mathcal{C}$$, with for all objects $$C$$ in $$\mathcal{C}$$ and $$D$$ in $$\mathcal{D}$$: $$\mathcal{D}(F(C), D) \cong \mathcal{C}(C, G(D)).$$ That is, $$F$$ is left adjoint to $$G$$. Then taking the dual everywhere, this is precisely the same as $$\mathcal{D^\mathrm{op}}(D, F^\mathrm{op}(C)) \cong \mathcal{C^\mathrm{op}}(G^\mathrm{op}(D), C).$$ So indeed we see that $$F^\mathrm{op}$$ is right adjoint to $$G^\mathrm{op}$$.
So to sum up, colimits in $$\mathcal{C}$$ are the same as limits in $$\mathcal{C}^\mathrm{op}$$. If $$F: \mathcal{C} \to \mathcal{D}$$ is a left adjoint, then $$F^\mathrm{op}: \mathcal{C}^\mathrm{op} \to \mathcal{D}^\mathrm{op}$$ is a right adjoint. So since right adjoints preserve limits, $$F^\mathrm{op}$$ preserves limits and thus $$F$$ preserves colimits.
• As far as I know, if $A,B\in Ob(\mathcal A)$, then $\mathcal A^{op}(A,B)=\mathcal A(B,A)$. So shouldn't we have $\mathcal D^{op}(F(C),D)=\mathcal D(D,F(C))$ then? How does the equality $\mathcal D^{op}(F(C),D)=\mathcal D(D,F^{op}(C))$ agree with the first formula in my comment? Also, I'm not very familiar with the "opposite functor". I suppose $F^{op}$ is defined on object in the same way as $F$, right? How is it defined on morphisms? Jun 25 '19 at 22:06
• @user634426 You are right that $\mathcal{A}^\mathrm{op}(A, B) =$\mathcal{A}(B, A)$. That is exactly what I applied to both sides of the equation. Look again carefully ;) Jun 25 '19 at 22:16 • @user634426 About the dual functor:$F^\mathrm{op}$does the same as$F$but on the dual categories. So it is indeed defined in the same way on objects as$F$. For arrows: given an arrow$f: A \to B$in$\mathcal{C}$, let us denote by$f^\mathrm{op}: B \to A$the corresponding arrow in$\mathcal{C}^\mathrm{op}$(similar for arrows in$\mathcal{D}$). Then$F^\mathrm{op}(f^\mathrm{op}) = F(f)^\mathrm{op}$. So we take the original arrow, see where it is sent by$F$and then reverse it again. Jun 25 '19 at 22:17 • hmm, applying$\mathcal A^{op}(A,B)=\mathcal A(B,A)$to the LHS (i.e., to$\mathcal D(F(C),D)$) -- in which case$\mathcal A=\mathcal D, A=F(C), B=D$-- would yield, as far as I can see,$\mathcal D(D,F(C))$and not$\mathcal D(D,F^{op}(C))$. I must be not seeing something or looking at the wrong place... Jun 25 '19 at 22:25 • @user634426 That is because you have$\mathcal{A}^\mathrm{op}$on the left side of your equality sign, while I start with$\mathcal{D}$(and not$\mathcal{D}^\mathrm{op}$. To apply$\mathcal{A}^\mathrm{op}(A, B) = \mathcal{A}(B, A)$you have to set$\mathcal{A} = \mathcal{D}$(like you did) and$B = F(C)$and$A = D\$ (these you mixed up). Jun 25 '19 at 22:27