Proof that the set of infinitesimal does not consist only of the zero element in Synthetic Differential Geometry Thanks to Z. A. K. answer I edited my question
In this lecture Synthetic Differential Geometry they have the following
Definition 4.3. An infinitesimal on $R$ is any nilsquare element of $R$, i.e. $x^{2}=0 .$ We denote the collection of infinitesimals on $R$ by $\Delta:=\left\{x \in R \mid x^{2}=0\right\}$.
Axiom 4.1. (Kock-Lawvere) For any mapping $g: \Delta \longrightarrow R$ there exists a unique $b$ in $R$ such that for all $\varepsilon$ in $\Delta$ we have $g(\varepsilon)=g(0)+b \varepsilon$.
Theorem 4.2. $
 \Delta \neq\{0\} \text {. }
$
the proof of the theorem goes like this :
$g(\varepsilon)=g(0)+\varepsilon b$ for $b=0$, since $g(\varepsilon)=g(0)=0=0+\varepsilon 0$, but also for $b=1$, since $g(\varepsilon)=g(0)+\varepsilon=\varepsilon$. So this $b$ is not unique (because $0 \neq 1$ ), which contradicts the Kock-Lawvere axiom. So $\Delta$ cannot coincide with $\{0\}$.
Why is that  if $\Delta \neq\{0\}$ would solve the contradiction in the proof?
 A: Let me restate the axiom for you.

Kock-Lawvere axiom: for any map $g: \Delta \rightarrow R$, we can find precisely one element $b_g \in R$ that satisfies the following:

*

*for every $\varepsilon \in \Delta$, the equality $g(\varepsilon) = g(0) + b_g\varepsilon$ holds.


What happens if we assume $\Delta = \{0\}$? Well, under that assumption we would know that

*

*Every $\varepsilon \in \Delta$ satisfies $\varepsilon = 0$.

*Accordingly, for any map $g: \Delta \rightarrow R$ and any $\varepsilon \in \Delta$, we have $g(\varepsilon) = g(0)$.

But this would mean that for any map $g: \Delta \rightarrow R$ we can in fact find two different numbers, $b_g=0$ and $b_g=1$, so that for every for every $\varepsilon \in \Delta$, the equality $g(\varepsilon) = g(0) + b_g\varepsilon$ holds. If $b_g=0$ then we have equalities $g(\varepsilon) = g(0) + 0 \varepsilon = g(0)$, if $b_g=1$, we would also have $g(\varepsilon) = g(0) + 1 \varepsilon = g(0) + 1 \times 0 = g(0)$. This contradicts the Kock-Lawvere axiom, which asserts that we can find exactly one number with this property.
Since assuming $\Delta = \{0\}$ leads to a contradiction, our assumption must have been wrong, and so we can conclude $\Delta \neq \{0\}$.
edit: Note that if we do not assume $\Delta = \{0\}$, then neither of 1,2 above hold, and we cannot get a contradiction. Since it is not the case that $\Delta$ consists of only one element, it's perfectly fine and non-contradictory that $g(d) = g(0) + 0d$ and $g(d) = g(0) + 1d$ both work for some specific element $d \in \Delta$, as long as they don't both work for all elements $\varepsilon \in \Delta$, the K-L axiom is not violated.
