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Motivation: In May's Concise Course in Algebraic Topology, the colimit is brought up to talk about van Kampen. Here's the diagram (from wikipedia) for ease of notation:

enter image description here

He notes that $\phi_{X}, \phi_{Y}$ are the initial such morphisms into the colimit $L$. In most cases, it's clear that this means something like "an injection." Although,...


  1. Because the colimit is an object satisfying this diagram, how can we force $\phi_{X}$ and $\phi_{Y}$ to be initial if we are not a priori considering the colimit object $L$? It doesn't seem like we can take initial morphisms in a hom-set where the destination object is not fixed.
  2. Similar to the first part, when people talk about the colimit of some objects, they are implicitly specifying two "injection" mappings (in this case $\phi_{X}$ and $\phi_{Y}$); how do we pick these?

As an example for the second, if we consider the diagram for the product in the category of Groups and make the "projections" just zero mappings, then we will not obtain the same product as if we make them the "standard" projections --- I'm not entirely sure if we will get a product at all in that case.

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Think of it as L=X+Y.(It's not a product) Easy way to think is as if L object was split to two parts, and the morphisms would choose elements from the parts. – tp1 Apr 5 '12 at 19:47
up vote 1 down vote accepted

Let me answer your second question first: we don't "pick" $\phi_X$ and $\phi_Y$, they are part of the data of what it means to have a colimit. In other words, the colimit is not simply the object $L$, but rather it is the object $L$ together with the collection of maps $\phi_X$ satisfying the required properties. In your groups example, the product (in the category of groups) of two groups is not simply the cartesian product of the two groups, it is this cartesian product together with the projections onto the two factors; the cartesian product of two groups with "projections" zero mappings is not what we would refer to as the product of the two groups in the category of groups.

In general, when defining some object via a universal property, the morphisms involved are part of the definition. May's claim is that $L$ together with the collection of morphisms $\phi_X$ is the initial object in the category whose objects are pairs $(N,\{\psi_X\})$ where $N$ is an object in the original category and $\psi_X$ are morphisms making the "outside" diagram in the picture you gave commute.

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In the first part of your answer: my main concern was, what if there are two collections of maps $\{\phi_{X}\}$ and $\{\psi_{X}\}$ which satisfy the required properties? Are these necessarily related in some way? – james Apr 5 '12 at 20:27
In the second part: Is this kind of "initial" object requirement because we could take some larger element in the category that our $\phi_{X}$ do not map into and so are sort of "unnecessary" to have in the colimit? – james Apr 5 '12 at 20:31
@james: Yes, they are related: the relation comes straight from the definition of colimit. Just plug your $\psi_X$ into the diagram, and the map $u$ is the relation. Swap the roles of $\psi$ and $\phi$ and you get a different $u$ (the inverse of the first $u$) which reverses the relation. – Hurkyl Apr 5 '12 at 21:17

I am not sure if I understand your question. But I hope the following statements answer your question.

a) Even if we're sloppy and say that "$L$ is the colimit of this diagram", what we really should be saying is that "$L$, together with the maps $\phi_X$, $\phi_Y$, ... form a colimit". The maps are part of the data.

b) One should be careful with just what it is that is initial here. I think the statement you are after is that a colimit is an inital object in the category of co-cones. That is, there is a category whose objects are a choice of $L$ together with maps $\phi_X$, $\phi_Y$, ..., and whose morphisms are maps $u$ like in your diagram that make everything commute. This is the category of co-cones, and a colimit is an initial co-cone. NB: This does not necessarily mean that some map is an injection!

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Yes, (b) this is what I meant. I'm still a bit unsure if this is what May meant (see Santiago Canez's answer above) especially because I do not think he has defined co-cones yet. – james Apr 5 '12 at 20:32

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