In Algebra, we just finished chapter 10 of Gallian. One of the big theorems in the chapter is the First Isomorphism Theorem – and since I’m a lazy typist I’ll call it FIT from now on. (Isomorphism theorems 2 and 3 are left as exercises.) Skipping over what a group is, what a homomorphism is, what the kernel of said homomorphism is, and what normal subgroups and factors groups are (see, FIT talks about all sorts of cool stuff!), FIT says:

Suppose \theta :G \longrightarrow G^\prime is a homomorphism. Then G/\text{ker}\theta \cong G^\prime\theta; moreover, the map \theta : G/\text{ker}\theta \longrightarrow G^\prime\theta by g\text{ker}\theta \mapsto g\theta is an isomorphism.

One of the points of the proof, since \theta is defined on the cosets of G/\text{ker}\theta by specifying the image of a representative, is whether or not the map \theta is even well defined for some g\text{ker}\theta \in G/\text{ker}\theta. In particular, will taking different representatives of the coset g\text{ker}\theta always have the same image? It turns out that the map is indeed well defined – it’s a rather straightforward argument – but this just brings up another question:

If a map is well defined, does it have the property of well-definededness?

Yay, English!

EDIT: Fixed my accidental interspersing of \varphi for \theta.

This entry was posted in Algebra, Groups, Math, Off Topic. Bookmark the permalink.

One Response to Well-Definededness

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