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Let $\mathrm{vect}$ be the category of vector spaces over $k$ and let $V \in \mathrm{vect}$. Can one say $\mathrm{Hom}(V, V)$ (the hom-space in the category $\mathrm{vect}$) is isomorphic to $k$?

Thanks in advance!

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    $\begingroup$ Certainly not for all $V$... $\endgroup$ – Arnaud D. Feb 25 at 11:37
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Say $V$ is finite dimensional so $V\cong \oplus_{i=1}^n k_i$ where $k_i=k$. Then

$$\mathrm{Hom}(V,V)\cong \mathrm{Hom}(\oplus_{i=1}^n k_i, \oplus_{j=1}^n k_j)=\oplus_{j=1}^n \mathrm{Hom}(\oplus_{i=1}^n k_i,k_j)=$$ $$\oplus_{i=1}^n \oplus_{j=1}^n \mathrm{Hom}(k_i,k_j) \cong \oplus_{i=1}^n \oplus_{j=1}^n\mathrm{Hom}(k,k)\cong \oplus_{i=1}^n \oplus_{j=1}^n k,$$

which is not $k$ in general, but finitely many copies of it. In the infinite dimensional case this is even further from being true. Your claim is of course true when $\dim(V)=1$.

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  • $\begingroup$ Thanks for you answer! In the answer to my question math.stackexchange.com/questions/3118727/…, the person answering uses $\mathrm{Hom}_{\mathcal{M}}(M, M) = k$ where $\mathcal{M}= vect$ the module category of vector spaces. Do you have an idea why it holds there? I remember my professor saying something similar, so I suppose that the claim itself is correct. $\endgroup$ – P. Schulze Feb 25 at 11:49
  • $\begingroup$ I have in my case a module category $\mathcal{M}$ over a category $\mathcal{C}$ and $M$ is irreducible. Does irreducible mean that $M$ has dimension 1? $\endgroup$ – P. Schulze Feb 25 at 11:50
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    $\begingroup$ In the linked post you are overlooking the adjective “irreducible.” Vector spaces of dimension greater than 1 are not irreducible. $\endgroup$ – Randall Feb 25 at 11:50
  • $\begingroup$ @Randall thanks! $\endgroup$ – P. Schulze Feb 25 at 11:52
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    $\begingroup$ I want to add a reference to @Randall comment $\endgroup$ – Javi Feb 25 at 11:53

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