What does "$\cong$" sign represent? - Mathematics Stack Exchange
In geometry, $\cong$ means congruence of figures, which means the figures have the same shape and size. (In advanced geometry, it means one is the image of the other under a …
Proof of $(\\mathbb{Z}/m\\mathbb{Z}) \\otimes_\\mathbb{Z} …
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Difference between "≈", "≃", and "≅" - Mathematics Stack Exchange
$\cong$ is used to show a congruency between two mathematical expressions, which could be geometrical, topological, and when using modulo arithmetic you can get different numbers that …
Computing the Canonical bundle $K_{\\mathbb{P}^n} \\cong …
2023年8月22日 · Q1: Yes, this is the definition of the determinant of a one-dimensional vector space. Q2: Yes, the dual of the trivial line bundle is the trivial line bundle (for instance, use that …
Show that $\\operatorname{Hom}_{R} (R,M) \\cong M$
2017年5月28日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
abstract algebra - Why does $ (A/I)\otimes_R (B/J)\cong …
$\cong \hom((A \otimes_R B)/(I \otimes 1 + 1 \otimes J),T)$ By the Yoneda lemma, we are done. Remark: This is one of the thousands of trivial isomorphisms in basic algebra which are …
Connected sum of projective plane $\\cong$ Klein bottle
2014年11月28日 · How can I see that the connected sum $\\mathbb{P}^2 \\# \\mathbb{P}^2$ of the projective plane is homeomorphic to the Klein bottle? I'm not necessarily looking for an …
abstract algebra - Prove or disprove that $G_1/H_1 \cong G_2/H_2 ...
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group theory - how to show $SU(2)/\mathbb{Z}_2\cong SO(3 ...
$\begingroup$ This is an old question which already has an accepted answer which seems to be pretty clear. I'm not sure what the link that you give above adds to the conversation, and the …
What’s the difference between equals signs ≈, ≅, and ≃?
2017年6月29日 · I suppose similarity is an equivalence relation. $\equiv$ is "equivalent to" and appears in regularly in modular arithmetic. $\approx$ is aproximately. $\cong$ is congruence …