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 $\overline{y}\equiv y'=y$.)

  • $\begingroup$ 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. $\endgroup$ Jan 29, 2017 at 10:53
  • $\begingroup$ Correction: I meant the other side of the diagram: $F(B)$ and two copies of $1_{F(B)}$. $\endgroup$ 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
  • 1
    $\begingroup$ $y$ 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. $\endgroup$ 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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.