Thanks again for bringing them to our attention.

]]>The often recommended Catsters are highly over-rated though much better than nothing. (I like the people, but the presentation & videos have numerous problems.)

Steve Roman’s introduction (see above) is quite good but not nearly as long or complete (seems to have stopped after 6 lessons.)

Codrington’s (also above) is quite interesting and a unique presentation but also stopped after a bare introduction.

Awodey is good, but also must less complete.

]]>This week, a few of the regulars had emailed me over the weekend, or early this morning saying they were travelling or had work conflicts, so I had let this evening’s meeting slide. I thought you had been on the email I sent out to that effect, but perhaps I’d missed you since you’d only joined us in the last few weeks. Again, my apologies.

I’ll try to get us back on schedule with official invitations for next Monday evening. Thanks for sticking with us!

]]>In fact you are correct, U is exactly a functor if Q is an actual category. Of course but the remark states that the existence of U guarantees that Q is a category. ]]>

What $A$ shows is that $U$ is a function.

What $B$ shows is that $U$ preserves morphisms and that too injectively which becomes quite handy when we want to prove (1,2).

As an example, take Groups. Group elements can be represented as invertible matrices which form a subgroup of General Linear Group -. $GL(V)$. Thus, we have group morphism or homomorphism and a map.

Then a function similar to $U$ called $\rho$ maps each group element $g$ to $\rho(g)$. And for $g_1,g_2 \in G$, we have matrices $\rho(g_1),\rho(g2)$. Since $g_3=g_1*g_2$, to preserve group structure, we need $\rho(g_1*g_2) = \rho(g_1)*\rho(g_2)=\rho(g_3)$. This is same as $B$ but also includes additional requirement – injection.

How does the injection help?

Take questionable Category $Q$. Then attempt to prove $1$ which is identity morphism in the domain and co-domain.We need to prove $f \circ id_x = f$ and $id_y \circ f = f$. To prove $f \circ id_x = f$ in $Q$, we need to move this equation to known category $C$ using $U$.

Thus,

$U(f \circ id_x) = U(f) \circ U((id_x)

= U(f) \circ id_{U(x)}

= U(f)$

Because of injectivity, we can assert $f \circ id_x = f$.

]]>Homomorphism of the groups preserve group operation for a mapping between groups. In similar fashion, morphisms preserve operations between objects of a category for mappings between objects of a given category. ]]>