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If $g \circ f$ is monic, then $f$ is monic. Or, if $g \circ f \circ h = g \circ f \circ k \implies h = k$, then $ f \circ h = f \circ k \implies h=k$.

I am not exactly sure how to prove this. I don't know what my possible actions or manipulations are to get from one statement to the other. I drew a (messy) commuting diagram (no idea how to do it in LaTeX), where everything commutes, but I don't know how to turn that into a proof.

The commuting diagram looked like this: a square with h on top, k on the left, and $g\circ f$ on the bottom and right - this square commutes by the assumption - then I added in the composition triangles for each $g\circ f$, identifying the codomain of f in the middle of the square, making it look like a pushout diagram. This makes a new square with h on top, k on the left, and f in the bottom and right, the commuting diagram for a monomorphism. Everything else in the diagram commutes. So I guess a second question, does that mean the square commutes? (A diagram commutes if every triangle in it commutes, is the reverse true?)

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Write down the proof in the category of sets. Then replace elements and functions by morphisms. Done. Or, actually the general case can be reduced to this (Yoneda). – Martin Brandenburg Jan 10 '13 at 23:37
up vote 6 down vote accepted

You don't need to draw any commutative diagrams for this problem (but if you ever do you can learn quite quickly how to draw very complicated diagrams using XYpic (works with Latex and/or Lyx)).

Given that $g\circ f$ is monic we show that $f$ is monic. Assume that $f\circ h = f\circ k$ holds. The claim will follow by showing that $h=k$. Apply $g$ on the left side of the equality $f\circ h = f\circ k$ to obtain $g\circ (f\circ h)= g\circ (f\circ k)$. Using associativity of the composition we obtain $(g\circ f)\circ h = (g\circ f)\circ k$. But, $g\circ f$ is monic (so left cancelable) and so $h=k$.

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I know, I just like commutative diagrams, and I just wanted to see what it looks like, and it was another option. – fhyve Jan 11 '13 at 5:31

There's no need for a commutative diagram to prove this:

Using the fact that $g\circ f$ is monic, we show that $f$ is monic.

Assume $$f\circ h = f\circ k\tag{1}$$

Compose $g$ on the left with each of $f\circ h, f\circ k$ in equation $(1)$:

$$g\circ (f\circ h)= g\circ (f\circ k).\tag{2}$$

By associativity of composition, $(2)$ is equivalent to

$$(g\circ f)\circ h = (g\circ f)\circ k.\tag{3}$$

But we are given $g\circ f$ is monic, so from $(3)$, we have that $$ h=k\tag{4}$$

Hence we have shown that $$f\circ h = f\circ k \implies h = k$$

and thus, $f$ is monic.

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You had an extra ‘$’ in the middle. :) – Haskell Curry Jan 10 '13 at 23:52

I think a diadram would have said all that and saved a lot of space, and a lot of time. But im sure u feel smarter for it. Some people learn in different ways. If u know something ur goal is to communicate that thing in a clear and concise manner, elagance is not a prerequisite for understanding. The requisite that f is monic is exactly that gf is monic since both arrows f and gf are a requisite to the universal morphism g, in particular f and gf are both co-equalizers of any parallel pair of arrows equal on the domain of f, which is immediate from the diagram. There is no algebra needed to solve this problem, only a diagram and a understanding of the universal mapping properties. 99% of the proof is in your head. And to answer your original question the diagram is equivalent to the diagram of a co-equalizer with f and gf are the two co-equalizers and g the limit of f and gf.

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You are claiming in particular that $g$ is uniquely determined by $f$ and the composite $gf$ provided both are monic. This is very far from true... – Martin Mar 12 '13 at 5:31
Amazing answer..... – nicolas Mar 21 '14 at 10:51

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