Well spotted. The mistake in the ’proof’ is in confusing cones over $(d \to e \to f, c \to f)$ and cones over $(d \to e, e \to f, c \to f)$. I think they are the same iff the right square is a pullback.

Edit: I made the same mistake at comma object. Both fixed now, I think.

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A --> B --> C

| | |

v v v

D --> E --> F

is a pullback then the two statements

(i) the outer rectangle is a pullback

(ii) the left square is a pullback

are equivalent. But it may well happen that both (i) and (ii) hold,

without the right square being a pullback.

For instance let i: A --> B be a split mono with retraction p: B --> A

and consider

A ==== A ==== A

| | |

| |i |

| v |

A ---> B ---> A

i p

where all unnamed maps are identity maps. Then the left square and

the outer rectangle are pullbacks but the right square cannot be

a pullback unless i was already an isomorphism. ]]>

Added a proof of the pasting lemma to pullback, and the corresponding lemma to comma object (also added the construction by pullbacks and cotensors there).

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