A question here concerns much the same issue, but I don't understand the explanation there, so I will ask here for a more detailed explanation.
Excerpt:
Discussion:
I think I understand the discussion up to "essentially the same". I think I understand why "essentially the same" implies the diagram as described, but I don't understand why the vertical homomorphisms are isomorphisms.
Here's what I understand about the diagram so far. The middle vertical line is obviously the identity function. Regarding the left vertical line, call it $l$, and labeling the arrow $H \to G$ as $j$ for inclusion, we want $j \circ l = \text{id} \circ f$.
Regarding the third vertical line, call it $r$, and labeling the arrow $G \to G/H$ as $\varphi$ for the canonical function sending group elements to their cosets, we want $\varphi \circ \text{id} = r \circ g$.
How do I know that $l$ and $r$ are isomorphisms? I understand that for both sets of horizontal arrows, the first one is the homomorphism from identity to identity, the second one has to be injective (because only the identity can be sent to the identity for the second arrows), and the third arrow has to be surjective because the fourth arrow sends everything to the identity.
I appreciate any help.
Edit:
I think I can see why the right vertical arrow is an isomorphism. It follows pretty quickly from the First Isomorphism Theorem, as Lee Mosher said. I can't figure out why the left vertical arrow is an isomorphism though.
Edit 2:
I think I've made some progress with the right side, so I'll type that out here. Consider the map $\lambda \colon G/H \to G''$ given by $xH \mapsto g(x)$. We show that this is an isomorphism.
First, note that $\lambda (xH yH) = \lambda (xyH) = g(xy) = g(x)g(y)$. On the other hand, $\lambda (xH) \lambda (yH) = g(x) g(y)$, so $\lambda$ is a homomorphism.
To show injectivity, we show that $\lambda$ has trivial kernel. Consider $xH \in G/H$ such that $\lambda (xH) = g(x) = e'' \in G''$. Thus $x \in \text{ker } g$. Because of the way the cosets of $G/H$ partition $G$, we must have $x \in H$, hence $xH = H$.
To show surjectivity, consider $x'' \in G''$. We know there is some $x \in G$ such that $g(x) = x''$ because $g$ must be surjective, given the exact sequence. Thus $\lambda (xH) = g(x) = x''$, so we conclude that $\lambda$ is surjective and therefore an isomorphism.
The isomorphism going the other way, $\lambda^{-1} \colon G'' \to G/H$, will be given by $x'' \mapsto xH$, where $x$ is such that $g(x) = x''$.
Now we want to show that $\lambda^{-1} \circ g = \varphi \circ \text{id}$, i.e. that the right square commutes. Consider $x \in G$, where $g(x) = x'' \in G''$. We have $\lambda^{-1} \big(g(x) \big) = \lambda^{-1}(x'') = xH$. Going other way around the square, we have $\varphi \big(\text{id } x) \big) = \varphi (x) = xH$.
I think that's correct. Hopefully I can come back and describe the left square as well.
Edit 3:
For the left square, we need $\text{id } \circ f = j \circ l$. Given some $x' \in G'$, we have $\text{id } \circ f(x') = f(x') \in G$. We need $j \circ l (x')$ to match this. Since $j$ is just the inclusion of $H$ into $G$, we try defining $l \colon G' \to H$ as $x' \mapsto f(x')$. This is valid because $H = \text{ker } g = \text{im } f$.
According to this definition, $l$ is a homomorphism because $l(x' y') = f(x' y') = f(x') f(y') = l(x') l(y')$. It's injective because the kernel is trivial. To see this, consider $x' \in G'$ such that $l(x') = f(x') = e \in H$. Since $f$ is an injective homomorphism, we must have $x' = e' \in G'$, hence $l$ has a trivial kernel. To see that $l$ is surjective, consider $h \in H$. Because $H = \text{im } f$, we have $x' \in G'$ such that $f(x') = h$, which means $l(x') = h$. Thus $l$ is an isomorphism.
We show that the left square commutes. Consider some $x' \in G'$. We know that $\text{id } \circ f(x') = f(x') \in G$. Going the other way around the square, we have $j \circ l (x') = j \big(f(x')\big) = f(x')$, as desired.
I think this is correct. Thanks to Lee Mosher for helping me.
First of all, "essentially the same" is not a mathematical term, and it does not make sense to ask whether "essentially the same" implies the statement in the sentence to follow. "Essentially the same" is an intuitive term that the author uses in order to motivate the sentence to follow.
Of course, one still has to prove that sentence, but the proof is more or less an immediate application of the so-called First Isomorphism Theorem.