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Source. Section 6, printed p. 152 (published PDF), describes the bases and circuits of this construction and explicitly omits the general matroid verification. The following is a compilation-supplied expansion of that same construction.

Statement. Let $M=(E,\mathcal F)$ be a finite matroid, $e\in E$, and let $S$ be a disjoint new set of $k\ge1$ elements. On $E'=(E\setminus\{e\})\cup S$, declare $A\cup T$ independent, where $A\subseteq E\setminus\{e\}$ and $T\subseteq S$, exactly when

$$A\in\mathcal F,\qquad T\ne S\ \text{or}\ A\cup\{e\}\in\mathcal F. \tag{1}$$

This defines a matroid. Its bases are exactly

$$\begin{cases} (B\setminus\{e\})\cup S,& B\text{ a base of }M,\ e\in B,\\ B\cup(S\setminus\{s\}),& B\text{ a base of }M,\ e\notin B,\ s\in S. \end{cases} \tag{2}$$

Its circuits are the old circuits omitting $e$ and the sets $(C\setminus\{e\})\cup S$ for old circuits $C$ containing $e$. Thus the new elements are in series, and $k=1$ simply relabels $e$.

Proof. Condition (1) contains the empty set and is hereditary. We verify augmentation and then use its equivalence to the finite matroid axiom. Let $A\cup T$ and $B\cup U$ satisfy (1), with

$$|A|+|T|<|B|+|U|. \tag{3}$$

First suppose $T\ne S$. If $|B|>|A|$, augment $A$ by an element of $B\setminus A$ in $M$; the new-copy part is still proper in $S$, so this also augments $A\cup T$ under (1). If $|B|\le|A|$, then (3) gives $|U|>|T|$. Choose $s\in U\setminus T$. It can be added if $T\cup\{s\}\ne S$, and can also be added if $A\cup\{e\}$ is independent.

The only remaining subcase has $|T|=k-1$, $U=S$, and $A\cup\{e\}$ dependent. Inequality (3) and $|B|\le|A|$ now force $|B|=|A|$. Since $B\cup U$ satisfies (1), $B\cup\{e\}$ is independent and has size $|A|+1$. Augment $A$ by an element of $(B\cup\{e\})\setminus A$. The augmenting element cannot be $e$, which was assumed not to augment $A$. It therefore lies in $B\setminus A$, and augments $A\cup T$ under (1).

Now suppose $T=S$, so $A\cup\{e\}$ is independent. If $U=S$, then $B\cup\{e\}$ is independent and $|B|>|A|$. Augmentation between these two old independent sets adds an element of $B\setminus A$ to $A\cup\{e\}$, as required by (1). If $U\ne S$, (3) instead gives $|B|\ge|A|+2$. Augment the independent set $A\cup\{e\}$ by the larger independent set $B$. Again the added element is in $B\setminus A$ and preserves (1) with all copies present. This proves augmentation in every case and hence the matroid property.

For the bases, an independent set with fewer than $k-1$ copies can still take a copy. If exactly $k-1$ copies are present, maximality requires that $A$ cannot be augmented by an old element and cannot take $e$ in $M$. These conditions say exactly that $A$ is an old base omitting $e$. If all $k$ copies are present, maximality says that $A\cup\{e\}$ is an old base containing $e$. Conversely, each set in (2) is independent by (1), and any further addition would augment its old base. Thus (2) lists precisely all bases.

Finally consider a minimal dependent set $A\cup T$. If $A$ is dependent in $M$, minimality makes $T=\varnothing$ and $A$ an old circuit omitting $e$. Otherwise $A$ is independent, so dependence in (1) requires $T=S$ and $A\cup\{e\}$ dependent. Its circuit contains $e$; minimality then forces $A=C\setminus\{e\}$ for that circuit $C$. Conversely, a set of this latter form is dependent; deleting a copy makes $T$ proper, and deleting an old element makes the corresponding subset of $C$ independent. It is therefore a circuit. This proves the asserted circuit list, and the circuit characterization gives the series property. The proof includes an old loop or coloop. $\square$

This fills the paper's stated local omission. It is not attributed to an author-issued erratum or to a separate published proof.