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I need to show that left adjoints preserve epimorphisms. I want to follow the following idea: $f: A\longrightarrow B$ is epic if and only if the diagram: $$ \begin{array}{ccc} A & \overset{f}\longrightarrow & B \\ {\scriptstyle{f}}{\downarrow} & & {\downarrow}\scriptstyle{1_B} \\ B & \underset{1_B}{\longrightarrow} & B \end{array} $$ is a pushout. So I suppose that $A,B\in \mathcal{C}$, $f$ in $\mathcal{C}$ and that the diagram above is a pushout. I consider the functors $F:\mathcal{C}\longrightarrow\mathcal{D}$ and $G:\mathcal{D}\longrightarrow\mathcal{C}$ such that $F\dashv G$ and I want to show that the diagram $$ \begin{array}{ccc} F(A) & \overset{F(f)}\longrightarrow & F(B) \\ {\scriptstyle{F(f)}}{\downarrow} & & {\downarrow}\scriptstyle{F(1_B)} \\ F(B) & \underset{F(1_B)}{\longrightarrow} & F(B) \end{array} $$ is a pushout. Hence the conclusion will follow.

However, I don't think this way of reasoning is totally fine: it turns out that I don't use the hypothesis of $F$ being left adjoint to $G$ in the proof that the second diagram above is a pushout. I know it is really simple and basic, but I feel like I am not seeing something. Can anyone please help me?

(For the record, this is the sketch of my proof that the second diagram is a pushout. For sure it commutes ($F$ respects composition, being a functor). I consider an object $Y\in \mathcal{D}$ and maps such that the diagram $$ \begin{array}{ccc} F(A) & \overset{F(f)}\longrightarrow & F(B) \\ {\scriptstyle{F(f)}}{\downarrow} & & {\downarrow}\scriptstyle{y'} \\ F(B) & \underset{y}{\longrightarrow} & Y \end{array} $$ commutes. I need to find a unique $\overline{y}:F(B)\longrightarrow Y$ such that $\overline{y}\circ 1_{F(B)}=y'$. Take $\overline{y}\equiv y'=y$.)

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  • $\begingroup$ What is your proof that the second diagram is a pushout? Note, that it isn't enough to show that the second diagram commutes, you need to show that it is initial such diagram i.e. that $F(A)$ and two copies $F(f)$ satisfy the universal property of pushouts. $\endgroup$
    – Derek Elkins left SE
    Jan 29, 2017 at 10:53
  • $\begingroup$ Correction: I meant the other side of the diagram: $F(B)$ and two copies of $1_{F(B)}$. $\endgroup$
    – Derek Elkins left SE
    Jan 29, 2017 at 11:11
  • $\begingroup$ I edited the question, with the idea I have. $\endgroup$
    – any_one
    Jan 29, 2017 at 11:33
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    $\begingroup$ $y$ and $y'$ don't need to be the same, in general, in which case there clearly isn't a $\bar y$ equal to both $y$ and $y'$. You need to do something to show that they in fact must be the same. This would be true if $F(f)$ was an epimorphism, but that's what we're trying to show so another argument is needed. $\endgroup$
    – Derek Elkins left SE
    Jan 29, 2017 at 12:06
  • $\begingroup$ hahaha I see, how stupid I am...I need to use the canonical bijection $\mathcal{B}(F(B),Y)\cong\mathcal{A}(B,G(Y))$ so I can show that in the category $\mathcal{A}$ the images through this bijection of $y$ and $y'$ need to be the same. And I can conclude that consequently $y$ and $y'$ are the same as well!!! Thank you, I apologise for the stupid question hahaha! $\endgroup$
    – any_one
    Jan 29, 2017 at 13:50

1 Answer 1

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That the image diagram is a pushout follows from the general fact that left adjoints preserve all colimits.

Assuming you haven't proven that yet, you can compute

$$ \begin{align*} \mathcal{D}(F(\operatorname{colim} X_j), Y) &\cong \mathcal{C}(\operatorname{colim} X_j, GY) \\&\cong \lim \mathcal{C}(X_j, GY) \\&\cong \lim \mathcal{D}(FX_j, Y) \end{align*} $$

which shows that $F(\operatorname{colim} X_j) \cong \operatorname{colim} F(X_j)$.

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