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For two matrices $\textbf{S}$ and $\textbf{T}$, a proof of $\det(\textbf{ST})=\det(\textbf{S})\det(\textbf{T})$ is given below in the diagrammatic tensor notation.

Here $\det$ denotes the determinant.

Why can the antisymmetrizing bar be inserted in the middle because "there is already antisymmetry in the index lines"?

For an introduction to the notation, you can refer to Figures 12.17 and 12.18 below.

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Attention: This question had an accepted answer, but the answer contained only a picture, the link for which is now broken and the user who posted it has since been deleted. I have therefore deleted the answer so it is no longer marked as answered. KalEl, if you have the picture you got as an answer here, or can reconstruct the answer without the picture, please do post it in a new answer. – Zev Chonoles Jun 12 '12 at 23:31
Please see the answer I have posted. – pre-kidney Jan 30 '13 at 2:16

(see explanatory note if confused)

Penrose writes "The antisymmetrizing bar can be inserted in the middle term because there is already antisymmetry in the index lines that it crosses." in the caption for Figure 13.8

What he means is that "antisymmetrization" is an idempotent tensor operator. This, in turn, follows from the fact that the antisymmetrizer can be viewed as a projection onto the totally antisymmetric subspace, and the fact that projection operators are idempotent. (See Antisymmetrizer)

Recall that an operator $I$ is idempotent iff $I^2=I$. In other words, we can replace the single bar in the diagram with two bars. That is all.

Explanatory Note: I have not posted images because (a) they are under copyright and (b) they are not relevant to the crux of the question.

I looked at the google cache of this webpage and noticed that all the attached diagrams were screenshots from the book "The Road to Reality" by Roger Penrose, which I have access to.

Figures 12.17 and 12.18 are located on pages 241 and 242, respectively.

The proof KalEl is referring to is found as Figure 13.8, on page 261.

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