About the $1's$ embracing a short exact sequence I'm studying the short exact among groups:
$$1 \xrightarrow{\alpha} H \xrightarrow{\beta} G \xrightarrow{\gamma} Q \xrightarrow{\delta} 1$$
It's clear to me $\beta$ is injective, $\gamma$ is surjective and $\text{Ker}(\gamma)=\text{Im}(\beta)$, leading to $Q \cong \frac G{\text{Im}(\beta)}$, where $\text{Im}(\beta) \cong H$, since injective.
What is not clear to me is the two $1$s embracing the sequence. I've read many different things about them.
First of all, $1$ seems to be used for groups only, representing some groups with just one element, i.e. their own identity.
So, $\alpha$ and $\beta$ are usually described as trivial homomorphism.
For $\delta$ is easier to me to understand it, since the trivial mapping which maps all $Q$ elements to the identity of the last "singleton" group.
Even I can understand it, it's not clear why this last $\delta$ is needed. Which kind of information is adding to the picture?
But for $\alpha$ this would map the identity of the first "singleton" group to all elements of H and really messes up my mind in doing that.
I've also read in another post which writing $1 \mapsto H$ implies $\beta$ is injective, but it's not clear to me how and why.
So, are them just "historical placeholders" or are adding informations?
Thanks
 A: We say that a sequence of morphisms $H\stackrel{f}{\to}G\stackrel{g}{\to}K$ is "exact at $G$" if and only if $\mathrm{Im}(f)=\ker(g)$.
(This is used mostly in groups, modules, and other abelian categories; less so in other categories where "kernel" may make no sense, like semigroups, or where it does not provide the same kind of information about the morphism, like monoids).
We say a longer exact sequence
$$\cdots \stackrel{f_{i-1}}{\longrightarrow}H_i \stackrel{f_{i}}{\longrightarrow}H_{i+1} \stackrel{f_{i+1}}{\longrightarrow}H_{i+2}\to\cdots$$
is "exact" (no qualifiers) if it is exact at each $H_k$.
So with a "short exact sequence"
$$1 \to H \stackrel{f}{\longrightarrow}G \stackrel{g}{\longrightarrow} K\to 1$$
we have a sequence that is exact, hence:

*

*Exact at $H$; which means that $\mathrm{ker}(f)$ is equal to the image of the unique map from the trivial group $1$ to $H$; i.e., $\mathrm{ker}(f)=\{e_H\}$, so $f$ is one-to-one.

*Exact at $G$; which means $\mathrm{Im}(f)=\ker(g)$.

*Exact at $K$; which means that $\mathrm{Im}(g)$ is equal to the kernel of the unique map from $K$ to the trivial group $1$; i.e., $\mathrm{Im}(g)=K$, so $g$ is surjective.

Thus, they are not placeholders, but provide the information that $f$ is one-to-one and $g$ is onto.
We use $1$s when we write the groups multiplicatively, but often use $0$s for abelian groups written additively. So you will sometimes see something like
$$ 0 \to \mathbb{Z} \stackrel{\cdot 2}{\longrightarrow} \mathbb{Z} \stackrel{\pi}{\longrightarrow} \mathbb{Z}/2\mathbb{Z}\to 0,$$
because the "trivial group" here is $\{0\}$.
Sometimes you'll see things like:
$$1 \to H \stackrel{f}{\longrightarrow} G\stackrel{g}{\longrightarrow} K$$
which tells you that $f$ is one-to-one, and that $\mathrm{Im}(f)=\mathrm{ker}(g)$, but with no assertion that $g$ is onto. E.g., in this question, which also uses $0$s.
Or
$$ H \stackrel{f}{\longrightarrow} G \stackrel{g}{\longrightarrow} K \to 1$$
which tells you that $g$ is onto, that $\mathrm{Im}(f)=\ker(g)$, but does not assert that $f$ is one-to-one.
So you do need the two $1$s to convey the full information.
A: These $1$s definitely have meaning. Notation $1=\{\ast\}$ and $1_G$ is the neutral element of the group $G$.
$1\xrightarrow{\alpha}H$: From the group axioms you have $\alpha(\ast)=1_H$. There is no other choice. Since $\alpha$ is a map of the underlying sets, it cannot take on multiple values.
Exactness at $H$: since $\ker(\beta)=\text{im}(\alpha)=\{1_H\}$, this implies injectivity of $\beta$. You do not need to put it as an extra condition, it is implied by exactness, because of the $1$ on the left.
Corresponding things hold for the homomorphism $\delta$ and its surjectivity as a cosequence of exactness at $Q$ and the $1$ on the right.
