# Countably infinite composition of injective functions.

Just out of curiosity that came from a topology homework assignment where I had to show the composition of 3 injective functions was injective.

Suppose $f_i : A_i \mapsto A_{i+1}$ were injective where $i \in \mathbb{N}$. I know that the composition of n such functions, i.e. $\bigcirc_{i=1}^n f_i$, is injective. But what about $\bigcirc_{i=1}^\infty f_i$? Is this an injective function?

Also, is composition bounded to countability? That is, I think that the definition of composition limits us to an ordering which means that we can't have a composition of uncountably many functions. Is this true?

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How do you define $\bigcirc_{i=1}^\infty f_i$? –  joriki Sep 10 '12 at 16:12
$\left( \cdots \circ f_n \circ f_{n-1} \circ \cdots \circ f_1 \right)$ –  torrho Sep 10 '12 at 16:17
Yes, I thought as much, but what does that mean? If $f_i:\mathbb R\to\mathbb R,x\to-x$, then what is $\bigcirc_{i=1}^\infty f_i$? –  joriki Sep 10 '12 at 16:19
oh wow. Thats almost analogous to the "chicken or the egg" delimma... So then, perhaps it is not possible to define deterministic infinite composition, correct? –  torrho Sep 10 '12 at 16:21
There is a useful notion of transfinite composition for an arbitrary number of functions. However, the construction is somewhat abstract (and sorry, I couldn't find a more palatable link). If you have functions between sets or topological spaces it turns out that the transfinite composition of injective functions still is injective. –  t.b. Sep 10 '12 at 17:47
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