Intuition about the first isomorphism theorem

Solution 1:

I understand the Theorem in the same way as you. The idea with a lot of these Algebra Theorems, where we factor an algebraic structure through a quotient, is to remove some undesirable part of the structure.

In this case, we want to get an isomorphism out of a surjective homomorphism, which is a much "nicer" map. So, we quotient out by the kernel, and as a result we have a map where only the zero element is sent to zero, which maintains surjectivity.

This sort of Theorem reappears frequently, and yours is the correct intuition.

Solution 2:

The fiber point of view is the one I like, because it captures the idea that when you quotient out $ker\phi $, you identify through $\sim $ all the points that $\phi $ sends to $0$.

To extend this a bit, suppose we take a topological space $X$, a space $Y$ and an $f:X\to Y$, and topologize $Y$ by declaring that $V$ open in $Y$ $\Leftrightarrow f^{-1}(V)$ open in $X.$

Now given any $\sim $ on $X$, $q:X\to X/\sim $ induces a topology on $X/\sim $ as above and from this you get the following result:

if $g:X\to Z$ is continuous, then there is a unique $\ \overline g:X/\sim \to Z$ such that $\overline g\circ q=g$

and so we have a topological analog of the First Isomorphism Theorem.