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.


This follows from duality. In general right adjoints preserve limits, and so by duality left adjoints preserve colimits. The fact that colimits are the dual of limits (and the initial object is the dual of the terminal object) should be clear.

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.

  • $\begingroup$ 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? $\endgroup$
    – user634426
    Jun 25 '19 at 22:06
  • $\begingroup$ @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 ;) $\endgroup$ Jun 25 '19 at 22:16
  • $\begingroup$ @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. $\endgroup$ Jun 25 '19 at 22:17
  • $\begingroup$ 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... $\endgroup$
    – user634426
    Jun 25 '19 at 22:25
  • $\begingroup$ @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). $\endgroup$ Jun 25 '19 at 22:27

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