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Edited to remove latex.mathoverflow invocation. I used the \mathsf font family so that it doesn't look too jarring compared to the accepted answer.
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Willie Wong
Willie Wong
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It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors T: Groups \rightarrow http://latex.mathoverflow.net/png?%5Crightarrow Groups$\mathsf{T}: \mathsf{Groups} \to \mathsf{Groups}$ with object function T(G) = G$\mathsf{T}(\mathsf{G}) = \mathsf{G}$ for every group G$\mathsf{G}$. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?

If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are not two such functors?

It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors T: Groups \rightarrow http://latex.mathoverflow.net/png?%5Crightarrow Groups with object function T(G) = G for every group G. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?

If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are not two such functors?

It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors $\mathsf{T}: \mathsf{Groups} \to \mathsf{Groups}$ with object function $\mathsf{T}(\mathsf{G}) = \mathsf{G}$ for every group $\mathsf{G}$. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?

If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are not two such functors?

added 134 characters in body
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Steven Gubkin
Steven Gubkin
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It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors T: Groups \rightarrow http://latex.mathoverflow.net/png?%5Crightarrow Groups with object function T(G) = G for every group G. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?

If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are not two such functors?

It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors T: Groups \rightarrow http://latex.mathoverflow.net/png?%5Crightarrow Groups with object function T(G) = G for every group G. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?

It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors T: Groups \rightarrow http://latex.mathoverflow.net/png?%5Crightarrow Groups with object function T(G) = G for every group G. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?

If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are not two such functors?

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Steven Gubkin
Steven Gubkin
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Two functors from Grp to Grp?

It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors T: Groups \rightarrow http://latex.mathoverflow.net/png?%5Crightarrow Groups with object function T(G) = G for every group G. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?